L'hospital rule for two variable.

There is no L'Hopital's Rule for multiple variable limits. For calculating limits in multiple variables, you need to consider every possible path of approach of limits.

What you can do here:

Put $x=r\cos\theta$ and $y=r\sin\theta$, (polar coordinate system) and $(x,y)\to (0,0)$ gives you the limits $r\to 0$ and no limits on $\theta$.

Now we need to substitute these in your problem which then becomes $$\lim_{r\to 0}\dfrac{r^2}{r(\cos\theta+\sin\theta)}=\lim_{r\to 0}\dfrac{r}{(\cos\theta+\sin\theta)}$$ Now for paths where $\cos\theta\to-\sin\theta$ or $\cos\theta=-\sin\theta$, the denominator $\to0$ or $=0$ while the numerator is just tending to $0$ but not exactly $0$.Since the second case exists $\implies$ the limit doesn't exist.

If extra condition that $(x+y)\neq 0$ is applied, even then the limit does not exist as for path $\theta=0$, limit is $0$ while for path $\sin\theta=-\cos\theta+r\cos^2\theta$, limit is something which can be non-zero.

The answer by Christian Blatter is correct. This particular limit is undefined.

I have recently written a paper giving a l'Hospital's rule for many types of multivariable limits. You can find it at

http://arxiv.org/abs/1209.0363

The limit doesn't exist, because every neighborhood $U_\epsilon({\bf 0})$ contains points $(x,y)$ where the expression $${x^2+y^2\over x+y}$$ is undefined.

Now one could be tempted to exclude the points $(x,y)$ with $x+y=0$ from consideration, but this is of no help: Let $S$ be the plane minus these points and consider an arbitrary sequence $x_n\to 0$ $\,(n\to\infty)$, $\, x_n>0$. Choosing $y_n:=-x_n+x_n^2$ ensures $(x_n,y_n)\in S$ and $(x_n,y_n)\to0$ $\,(n\to\infty)$. For such a sequence $${x_n^2+y_n^2 \over x_n+y_n}={2x_n^2+2x_n^3+x_n^4\over x_n^2}\to 2\qquad(n\to\infty)\ ,$$ whereas for the sequence $(x_n,0)$ the corresponding limit is $=0$.