By definition, 

$\displaystyle \vec{\bigtriangledown } \cdot \vec{f} = \frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}$, where $\displaystyle f_x, f_y, f_z$ are the respective $\displaystyle x,y,z$ components of $\displaystyle \vec{f}$.

Therefore, we need to calculate the above terms, shown as

$\displaystyle \begin{align*} \frac{\partial f_x}{\partial x}&=1 \\ \frac{\partial f_y}{\partial y}&= 0\\ \frac{\partial f_z}{\partial z}&= \frac{xy}{2\sqrt{z}} \end{align*}$

Therefore,

$\displaystyle R = 1 +\frac{xy}{2\sqrt{z}}$.

The gradient vector of f, $\displaystyle \nabla f$, is equal to $\displaystyle \left \langle f_x, f_y, f_z\right \rangle$. 

So, we must find the partial derivatives of the function with respect to x, y, and z, keeping the other variables constant for each partial derivative:

$\displaystyle f_x=yz+z\sec^2(xz)$

$\displaystyle f_y=xz$

$\displaystyle f_z=xy+x\sec^2(xz)$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

Plugging this in to a vector, we get

$\displaystyle \left \langle yz+z\sec^2(xz), xz, xy+x\sec^2(xz)\right \rangle$

The gradient of a function is given by

$\displaystyle \nabla f= \left \langle f_x, f_y, f_z \right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

Now, we find the partial derivatives:

$\displaystyle f_x=-ye^z\sin(xy)+yz$

$\displaystyle f_y=-xe^z\sin(xy)+xz$

$\displaystyle f_z=2z+e^z\cos(xy)+xy$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

To start, we must find the partial derivative with respect to y:

$\displaystyle f_y=2xyz^2$

The following rules were used to find the derivative:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we find the derivative of the above function above with respect to x:

$\displaystyle f_{yx}=2yz^2$

The rule used is stated above.

Finally, find the partial derivative of the above function with respect to z:

$\displaystyle f_{yxz}=4yz$

The rule used is already stated above.

The gradient of a function is given by

$\displaystyle \nabla f=\left \langle f_x, f_y, f_z\right \rangle$

So, we must find the partial derivatives. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

$\displaystyle f_x=z^2e^x$

$\displaystyle f_y=-3\sin(3y)$

$\displaystyle f_z=2ze^x$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

The gradient of a function is given by

$\displaystyle \nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are 

$\displaystyle f_x=y\sec^2(xy)$

$\displaystyle f_y=x\sec^2(xy)$

$\displaystyle f_z=2z+4$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

The gradient of a function is given by

$\displaystyle \nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$\displaystyle f_x=z\sec^2(xz)e^{\tan(xz)}$

$\displaystyle f_y=\cos(y)$

$\displaystyle f_z=x\sec^2(xz)e^{\tan(xz)}$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$. $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

The gradient of a function is given by

$\displaystyle \nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$\displaystyle f_x=2x$

$\displaystyle f_y=4y^3\cos(z)$

$\displaystyle f_z=4z^3-y^4\sin(z)$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

The gradient of a function is given by

$\displaystyle \nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$\displaystyle f_x=\sec(x)\tan(x)e^{\sec(x)}$

$\displaystyle f_y=z\sec^2(yz)$

$\displaystyle f_z=y\sec^2(yz)$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$

The gradient of a function is given by

$\displaystyle \bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants. 

The partial derivatives are

$\displaystyle f_x=z^4y+ze^{xz}$

$\displaystyle f_y=xz^4+2y$

$\displaystyle f_z=4z^3xy+xe^{xz}$

The partial derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$