$\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}$ doesn't exist (?) Is what I did correct?
We have: $\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}$
We know that as $x\rightarrow 0$ $\frac{\ln(1+x^3)}{x^3}=1$ , can be proven using taylor series..
So we have: $\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{x^3} \frac{x^4}{yx^2+y^3}$
Pose $y=Cx^2$ we get: $$\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{x^3} \frac{x^4}{Cx^4+C^3x^6}=\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{x^3} \frac{x^4}{x^4(C+C^3x^2)}=\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{x^3} \frac{1}{(C+C^3x^2)}=\frac{1}{(C+C^3x^2)}$$
Which is dependent on the constant $C$, Hence the limit doesn't exist.
What i'm doubtful about is the usage of $\lim_{x\rightarrow 0}\frac{\ln(1+x^3)}{x^3}=1$ in double variable limits, What i thought was since $x$ is still tending towards zero, and these terms are independent of y I can use it.
Thanks for your help! 

Yes you derivation is correct, indeed since $x\to 0$
$$\frac{\ln(1+x^3)}{x^3}\to1$$
it is true and we can conclude that for the trajectories $y=Cx^2\to 0$
$$\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}=\frac1C$$
then the limit doesn't exist.
