\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(14(0)^{3}y^{3} - 16(0)y^{5})}{(19(0)^{5}y + 24(0)^{6} + 5y^{6})}=\frac{0}{5y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{5y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(14x^{3}(x)^{3} - 16x(x)^{5})}{(19x^{5}(x) + 24x^{6} + 5(x)^{6})}=-\frac{1}{24}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(9(0)^{3}y^{3} + (0)^{5}y)}{(19(0)y^{5} + 3(0)^{6} + 18y^{6})}=\frac{0}{-18y^{6}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{-18y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(9x^{3}(x)^{3} + x^{5}(x))}{(19x(x)^{5} + 3x^{6} + 18(x)^{6})}=-\frac{1}{4}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

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

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(6(0)^{3} - 19(0)y^{2} + 13y^{3})}{(2(0)^{3} + 4y^{3})}=\frac{-13y^{3}}{4y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{-13y^{3}}{4y^{3}}=-\frac{13}{4}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(6x^{3} - 19x(x)^{2} + 13(x)^{3})}{(2x^{3} + 4(x)^{3})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{((0)y^{4} - (0)^{4}y)}{((0)^{5} + 4y^{5})}=\frac{0}{4y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{4y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(x(x)^{4} - x^{4}(x))}{(x^{5} + 4(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

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

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(14(0)^{3})}{(2(0)y^{2} - 4(0)^{3} + 4y^{3})}=\frac{0}{4y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(14x^{3})}{(2x(x)^{2} - 4x^{3} + 4(x)^{3})}=7\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(17(0)^{3}y^{2} + 3(0)^{4}y - 20y^{5})}{(3(0)^{5} + 4y^{5})}=\frac{20y^{5}}{4y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{20y^{5}}{4y^{5}}=5\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(17x^{3}(x)^{2} + 3x^{4}(x) - 20(x)^{5})}{(3x^{5} + 4(x)^{5})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(9(0)y^{5} - 2(0)^{6})}{((0)^{6} - (0)^{5}y + 8y^{6})}=\frac{0}{8y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{8y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(9x(x)^{5} - 2x^{6})}{(x^{6} - x^{5}(x) + 8(x)^{6})}=-\frac{7}{8}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(3(0)y^{3} - 13(0)^{3}y + 10(0)^{4})}{(5(0)^{4} + 4y^{4})}=\frac{0}{4y^{4}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{4y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(3x(x)^{3} - 13x^{3}(x) + 10x^{4})}{(5x^{4} + 4(x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)