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?
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:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n^4}-\frac{1}{n^2}}{\frac{1}{n}+\frac{1}{n^3}-\frac{1}{n}}=\lim_{n\to\infty} \frac{\frac{1}{n^4}-\frac{1}{n^2}}{\frac{1}{n^3}}=\lim_{n\to\infty} \frac{1-n^2}{n}= -\infty.$$
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 $x\not=0,y=0$

$\dfrac{xy}{x+y}=0\to0$