Does $\lim \frac{xy}{x+y}$ exist at $(0,0)$? Given the function $f(x,y) = \frac{xy}{x+y}$, after my analysis I concluded that the limit at $(0,0)$ does not exists.
In short, if we approach to $(0,0)$ through the parabola $y = -x^2 -x$ and $y = x^2 - x$ we find that $f(x,y)$ approaches to $1$ and $-1$ respectively. Therefore the limit does not exists.
I think my rationale is right. What do you think?
Alternatively, is there another approach for this problem?
 A: I'll explain here how to approach limits of functions in two variables, with the example the OP proposed in mind. If the limit
$$\lim_{(x,y)\to (0,0)} \frac{xy}{x+y}$$
exists and equals $L$, then it also follows that if $\{(x_n,y_n)\}$ is a sequence of points with limit $(0,0)$, then
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=L.$$
Now we can choose a number of easy sequences $\{(x_n,y_n)\}$ with limit $(0,0)$, and calculate the limit. For instance, we can pick points in a line $y=\lambda x$, with slope $\lambda$, i.e., $(x_n,y_n) = (\frac{1}{n}, \frac{\lambda}{n})$. In this case:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{\lambda}{n^2}}{\frac{1}{n}+\frac{\lambda}{n}}=\lim_{n\to\infty} \frac{\lambda}{(1+\lambda)n}$$
and the limit is $0$ as long as $\lambda\neq -1$. Hence, if the limit exists, it must be $0$. But the problem with $\lambda=-1$ tells us that there may be a problem if we approach $(0,0)$ with a path that ends tangent to $y=-x$ (notice that the function is not defined at points with $y=-x$).
Thus, next we look at a sequence following a path on a curve with tangent line $y=-x$ at $(0,0)$. Examples of such curves include $y=x^2-x$, $y=-x^2-x$ or $y=e^{-x}-1$. Thus, we may consider sequences $(x_n,y_n)$ given by:
$$\left(\frac{1}{n},\frac{1}{n^2}-\frac{1}{n}\right),\quad \text{or} \quad \left(\frac{1}{n},-\frac{1}{n^2}-\frac{1}{n}\right), \quad \text{or} \quad \left(\frac{1}{n},e^{-1/n}-1\right).$$
For the first sequence we obtain:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n}+\frac{1}{n^2}-\frac{1}{n}}=\lim_{n\to\infty} \frac{\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n^2}}=\lim_{n\to\infty} \frac{1-n}{n}= -1.$$
But the limit was supposed to be $L=0$. Hence the limit cannot exist. Similarly, if we try the other two sequences listed above:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{-\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n}-\frac{1}{n^2}-\frac{1}{n}}=\lim_{n\to\infty} \frac{-\frac{1}{n^3}-\frac{1}{n^2}}{-\frac{1}{n^2}}=\lim_{n\to\infty} \frac{1+n}{n}= 1,$$
and
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n}(e^{-1/n}-1)}{\frac{1}{n}+e^{-1/n}-1}=\lim_{n\to\infty} \frac{e^{-1/n}-1}{1+ne^{-1/n}-n}=0.$$
These results are inconsistent, and therefore the limit cannot exist. Even more dramatic: let $\{x_n,y_n\}$ be a sequence following the curve $y=x^3-x$ towards the origin, for instance put $(x_n,y_n)=(\frac{1}{n},\frac{1}{n^3}-\frac{1}{n})$. Then:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n^4}-\frac{1}{n^2}}{\frac{1}{n}+\frac{1}{n^3}-\frac{1}{n}}=\lim_{n\to\infty} \frac{\frac{1}{n^4}-\frac{1}{n^2}}{\frac{1}{n^3}}=\lim_{n\to\infty} \frac{1-n^2}{n}= -\infty.$$
A: The key point is to consider approaching the origin near the line $y = -x$.
No matter how small a neighborhood of the origin you consider, in that neighborhood $xy/(x+y)$ takes on every value. 
See the 
plot
 of $xy/(x+y)$.
You could also assume the limit exists and, using the definition of the limit of a multivariable function (with epsilons and deltas), arrive at a contradiction.
A: That is a good approach.  
Generalizing slightly, if $g(x)$ is a function defined in a punctured neighborhood of $0$ such that $\lim\limits_{x\to 0} g(x) = 0$ and $g(x)\neq 0$ for all $x$, then if the limit in question exists, you should have $\lim\limits_{(x,y)\to (0,0)}f(x,y)=\lim\limits_{x\to 0}f(x,g(x)-x)=\lim\limits_{x\to 0}x -\frac{x^2}{g(x)}$.  Such $g$ can be chosen to make this limit be any real number, $\infty$ or $-\infty$, or not exist in any sense.  Another particularly easy special path is to consider $\lim\limits_{x\to 0}f(x,0)=0$.
A: maybe I miss something but isn't it:
for $x=x,y=\dfrac{-x}{x+1}$ 
$\dfrac{xy}{x+y}=\dfrac{\dfrac{-x^2}{1+x}}{x-\dfrac{x}{1+x}}=\dfrac{\dfrac{-x^2}{x+1}}{\dfrac{x^2}{x+1}}=-1\to0$ when $(x,y)\to(0,0)$ 
and for $x\not=0,y=0$
$\dfrac{xy}{x+y}=0\to0$
