limit of $\frac{x^2}{x^2+y^2}$ To prove $\frac{x^2}{x^2+y^2}$ has no limit at $(0,0)$ we can take $y=kx$ and therefore:
$$\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2+y^2}=\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2+k^2x^2}=\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2(1+k^2)}=\lim_{(x,kx)\to (0,0)}\frac{1}{(1+k^2)}$$
For different $k$ we will get different limits, so there is no limit at $(0,0)$
But what about the limit at $(3,3)$ intuitively there is a limit which is $\frac{1}{2}$ but if we look at:
$$\lim_{(x,kx)\to (3,3)}\frac{x^2}{x^2+y^2}$$ we get
$$\lim_{(x,kx)\to (3,3)}\frac{1}{(1+k^2)}=?$$ 
Where am I getting it wrong?

The error is that you assume $y=kx\to3$, which is only true for $k=1$.  This is not the case for the limit to $3$, but it is true when $(x,y)\to(0,0)$.
