\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)^{3}y^{3} + (0)^{5}y - 3(0)^{6})}{(3(0)^{6} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(2x^{3}(x)^{3} + x^{5}(x) - 3x^{6})}{(3x^{6} + (x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(6(0)y^{2})}{(16(0)y^{2} + (0)^{3} - 4y^{3})}=\frac{0}{4y^{3}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(6x(x)^{2})}{(16x(x)^{2} + x^{3} - 4(x)^{3})}=\frac{6}{13}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(10(0)^{2}y^{3} + 15(0)^{4}y)}{(19(0)^{5} + 8y^{5})}=\frac{0}{8y^{5}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{8y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(10x^{2}(x)^{3} + 15x^{4}(x))}{(19x^{5} + 8(x)^{5})}=-\frac{25}{27}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{3}y + (0)^{4} + 19y^{4})}{(3(0)^{4} - 13(0)^{2}y^{2} + 4y^{4})}=\frac{19y^{4}}{4y^{4}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{19y^{4}}{4y^{4}}=\frac{19}{4}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(15x^{3}(x) + x^{4} + 19(x)^{4})}{(3x^{4} - 13x^{2}(x)^{2} + 4(x)^{4})}=-\frac{35}{6}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

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

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(8(0)^{6} + 19y^{6})}{(18(0)^{2}y^{4} - 12(0)^{4}y^{2} + 5(0)^{6} + y^{6})}=\frac{19y^{6}}{y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{19y^{6}}{y^{6}}=19\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(8x^{6} + 19(x)^{6})}{(18x^{2}(x)^{4} - 12x^{4}(x)^{2} + 5x^{6} + (x)^{6})}=\frac{9}{4}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(16y^{6})}{(2(0)^{5}y + (0)^{6} + 6y^{6})}=\frac{-16y^{6}}{6y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{-16y^{6}}{6y^{6}}=-\frac{8}{3}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(16(x)^{6})}{(2x^{5}(x) + x^{6} + 6(x)^{6})}=-\frac{16}{9}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

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

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

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)y^{5} - 3(0)^{3}y^{3} + (0)^{6})}{(2(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(2x(x)^{5} - 3x^{3}(x)^{3} + x^{6})}{(2x^{6} + 4(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)