Determine whether or not the limit exists: $\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$ 
Determine whether or not the limit $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$$
  exists. If it does, then calculate its value.

My attempt:
$$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \frac {2xy}{x^2+y^2} =\\&= 1 + \lim \frac 2{xy^{-1}+yx^{-1}} = 1+ 2\cdot\lim \frac 1{xy^{-1}+yx^{-1}}\end{align}$$
But $\lim_{x\to 0^+} x^{-1} = +\infty$ and $\lim_{x\to 0^-} x^{-1} = -\infty$
Likewise, $\lim_{y\to 0^+} y^{-1} = +\infty$ and $\lim_{y\to 0^-} y^{-1} = -\infty$
So the left hand and right hand limits cannot be equal, and therefore the limit does not exist. 

Your final answer is correct, but the way there is just slightly misleaded. The main problem is that you seperate your limit into a sum of limits. The first, $$\lim_{(x,y)\to(0,0)} \frac{x^2+y^2}{x^2+y^2}$$
converges and is equal to $1$, as you note. The second though, $$\lim_{(x,y)\to(0,0)} \frac {2xy}{x^2+y^2}$$
diverges.
The first problem (but probably not the most important) is that it doesn't diverge for the reasons you state (if you look closely, your argument actually concludes that the above limit is $0$). A proof of its divergence would be to consider the limit through points $(x,x)$ vs. through $(0,y)$, as suggested in another answer.
The more important problem though, is that this reasoning cannot conclude that the original limit doesn't exist! The equality (compactly stated) $$\lim f+g = \lim f + \lim g$$
holds only if both the limits of $f$ and of $g$ exist. In that case we can conclude that the original limit exists and is equal to blah blah. However there is nothing to be said in the case that either limit doesn't exist.
You could modify your method by keeping the expansions inside the limit operand: $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2} = \lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{x^2+y^2} + \frac{2xy}{x^2+y^2}$$
and conclude by the "path-dependence" argument.
Just to make clear what I mean, here's an example where the same argument would conclude that a very evidently convergent limit, doesn't exist!
$$0 = \lim_{x\to +\infty} \sin x-\sin x \stackrel{?}{=}\lim_{x\to +\infty}\sin x - \lim_{x\to +\infty}\sin x$$
