Is $f$ differentiable at $(0, 0)$? I asked a question earlier relating to the equation:
Let $f : \mathbb{R^2} \rightarrow \mathbb{R}$ be given by
$$f(x, y) = 
\cases{
    \frac{xy}{x+y}& if $x+y\neq0$  \\
    0 & if $x+y=0$}$$
I have another question relating to this equation which goes as follows:
Is $f$ differentiable at $(0, 0)$?
Im really not sure how to find if its differentiable or not so any help will be appreciated

The partial derivatives of $f$ at $(0,0)$ exist. They are equal to
$$ \partial f/\partial x(0,0) = 0, \ \ \ \partial f/ \partial y (0,0) = 0,$$
Therefore, if $f$ is to be differentiable at $(0,0)$, the derivative cannot be anything other than the linear transformation $ \mathbb R^2 \to \mathbb R$ represented by the matrix $[0 \ 0 ]$.
Let's test to see if the linear transformation $[0 \ 0]$ really is a derivative for $f$ at $(0,0)$. If $[0 \ 0]$ is the derivative, then, for any $\epsilon > 0$, there would need to exist a $\delta > 0$ such that
$$ \sqrt{h_1^2 + h_2^2} < \delta \implies \left|f(h_1, h_2) - f(0,0) - \left[0 \ 0\right] \left[ \begin{array}{c} h_1 \\ h_2\end{array}\right] \right| < \epsilon \sqrt{h_1^2 + h_2^2}$$
i.e. for any $\epsilon > 0$, there would need to exist a $\delta > 0$ such that
$$ h < \delta \implies \begin{cases}\left| h \frac{\cos \theta \sin \theta}{\cos \theta + \sin \theta} \right| < \epsilon h & {\rm  \ if \ } \theta \neq \frac {3\pi} 4,  \frac {7\pi}4 \\ 0 < \epsilon h & {\rm  \ if \ } \theta = \frac {3\pi} 4,  \frac {7\pi}4 \end{cases}$$
Here I introduced polar coordinates for $(h_1, h_2)$:
$$ h_1 = h \cos \theta, \ \ \ h_2 = h \sin \theta.$$
The problem is that, as $\theta$ approaches $ \frac {3\pi }4$ or $\frac {7\pi}4$, the expression $(\cos \theta \sin \theta)/(\cos \theta + \sin \theta)$ tends towards plus or minus infinity. (You can see this best if you draw a graph.) So for a given $ \epsilon$, the inequality
$$ \left| \frac{\cos \theta \sin \theta}{\cos \theta + \sin\theta} \right| < \epsilon$$
cannot be satisfied for all $\theta \in [0, 2\pi) \backslash \{ \frac {3\pi} 4,  \frac{7\pi }4 \}$.
Therefore, there exists no $\delta > 0$ such that 
$$ h < \delta \implies \left| h \frac{\cos \theta \sin \theta}{\cos \theta + \sin \theta} \right| < \epsilon h {\rm \ \ \ for \ all \ \  } \theta \in [0, 2\pi) \backslash \{ \tfrac {3\pi} 4,  \tfrac{7\pi } 4 \}.$$
The conclusion is that $f$ is not differentiable at $(0,0)$.

Alternative argument: If $f$ is differentiable at $(0,0)$, then its directional derivative at $(0,0)$ in the direction of the vector $(\lambda_1, \lambda_2)$ must be
$$ df/dt = \lambda_1 \partial f/\partial x + \lambda_2 \partial f/\partial y .$$
This is zero, since both partial derivatives vanish.
Now let's compute the directional derivative explicitly. Consider a straight path parametrised by $(x,y) = (\lambda_1 t , \lambda_2 t)$. Then
$$ f(t) = t \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$$
along this path (assuming that $\lambda_1 \neq \lambda_2$), so
$$ df/dt = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2},$$
which in general is NOT zero. This shows that $f$ is not differentiable at $(0,0)$.
