$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(13(0)^{3}y^{2} + 2(0)^{5})}{(14(0)^{2}y^{3} - 10(0)^{3}y^{2} - 3(0)^{5} + 15y^{5})}=\frac{0}{-15y^{5}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{-15y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(13x^{3}(x)^{2} + 2x^{5})}{(14x^{2}(x)^{3} - 10x^{3}(x)^{2} - 3x^{5} + 15(x)^{5})}=\frac{15}{16}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(5(0)^{3}y - 13(0)^{4} + 8y^{4})}{(3(0)^{4} + 2y^{4})}=\frac{-8y^{4}}{2y^{4}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{-8y^{4}}{2y^{4}}=-4\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(5x^{3}(x) - 13x^{4} + 8(x)^{4})}{(3x^{4} + 2(x)^{4})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{((0)^{2}y - 7(0)^{3})}{(6(0)^{3} + 9y^{3})}=\frac{0}{9y^{3}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{9y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(x^{2}(x) - 7x^{3})}{(6x^{3} + 9(x)^{3})}=\frac{2}{5}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(17(0)^{2}y^{2} + 14(0)y^{3} - 5y^{4})}{(8(0)^{3}y - 15(0)^{2}y^{2} + 3(0)^{4} + 2y^{4})}=\frac{-5y^{4}}{2y^{4}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{-5y^{4}}{2y^{4}}=-\frac{5}{2}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(17x^{2}(x)^{2} + 14x(x)^{3} - 5(x)^{4})}{(8x^{3}(x) - 15x^{2}(x)^{2} + 3x^{4} + 2(x)^{4})}=-13\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(16(0)y^{4} - (0)^{4}y)}{(24(0)^{5} - 17(0)^{3}y^{2} + y^{5})}=\frac{0}{y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(16x(x)^{4} - x^{4}(x))}{(24x^{5} - 17x^{3}(x)^{2} + (x)^{5})}=\frac{15}{8}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(18(0)^{4}y^{2} + 3y^{6})}{(6(0)^{5}y + (0)^{6} - 3y^{6})}=\frac{3y^{6}}{-3y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{3y^{6}}{-3y^{6}}=-1\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(18x^{4}(x)^{2} + 3(x)^{6})}{(6x^{5}(x) + x^{6} - 3(x)^{6})}=\frac{21}{4}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(16(0)^{3}y - 2(0)^{2}y^{2} + 16(0)^{4})}{(10(0)y^{3} + 20(0)^{3}y + 4(0)^{4} - 17y^{4})}=\frac{0}{-17y^{4}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{-17y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(16x^{3}(x) - 2x^{2}(x)^{2} + 16x^{4})}{(10x(x)^{3} + 20x^{3}(x) + 4x^{4} - 17(x)^{4})}=-\frac{30}{17}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(3(0)^{5}y - 15(0)^{2}y^{4} + 12y^{6})}{(4(0)^{6} + 2y^{6})}=\frac{-12y^{6}}{2y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{-12y^{6}}{2y^{6}}=-6\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(3x^{5}(x) - 15x^{2}(x)^{4} + 12(x)^{6})}{(4x^{6} + 2(x)^{6})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)^{4}y - 15(0)^{2}y^{3} + 7y^{5})}{(5(0)^{5} + y^{5})}=\frac{-7y^{5}}{y^{5}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{-7y^{5}}{y^{5}}=-7\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(8x^{4}(x) - 15x^{2}(x)^{3} + 7(x)^{5})}{(5x^{5} + (x)^{5})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{2}y^{3})}{(9(0)^{3}y^{2} - 16(0)^{2}y^{3} - 9(0)y^{4} + 2(0)^{5} + 2y^{5})}=\frac{0}{2y^{5}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{2y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(4x^{2}(x)^{3})}{(9x^{3}(x)^{2} - 16x^{2}(x)^{3} - 9x(x)^{4} + 2x^{5} + 2(x)^{5})}=\frac{1}{3}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$