Partial derivative of $ f(f(x,-x) , f(x,x)) $. 
$ f: \mathbb{R^2} \to \mathbb{R} $ is a function of $C^1$.
and $g: \mathbb{R} \to \mathbb{R}$ defined as :
$$ g(x) = f(f(x,-x) , f(x,x))$$

*

*Compute $g'(x)$.


*Suppose $f(0,0) = 0$. Determine $g'(0)$.


*Prove that if $f(0,0) = 0$ and $(0,0)$ is not a critical point of $f$ then, $g$ is stricly increasing near the point $(0,0)$.


I don't know how to answer question $3$, as I am not sure of my answers of question $1$ and $2$. Here is my attempt:

*

*Using the chain rule, I got:

$
\begin{align} g'(x) & = \frac{d}{dx} f(f(x,-x) , (x,x))\\
& =   \frac{df}{df}. \frac{df}{dx} (x,-x) + \frac{df}{df}. \frac{df}{dx} (x,x) \\
& = \frac{df}{dx} (x,-x) + \frac{df}{dx} (x,x)
\end{align}$


*$g'(0) = 2.\frac{df}{dx} (0,0)$
 A: $\frac{d}{dx}$ is used for functions which depend on the single variable $x$, and means (total) derivative with respect to that variable. $\frac{\partial}{\partial x}$ is used for functions which may depend on several variables, and means the (partial) derivative with respect to whichever of the several variables is called $x$ (usually the first one).
As per my first comment above, I think it's evil (intentional or not) of the problem author to let $g$ depend on $x$. I will therefore change the problem to be about
$$
g(t)=f(f(t,-t), f(t,t))
$$
Using the multivariable chain rule, we get (I hope I didn't mess anything up, this is a lot to keep straight)
$$
\frac{dg}{dt}(t)=\frac{d}{dt}f(t, -t)\cdot \frac{\partial f}{\partial x}(f(t, -t), f(t, t)) + \frac{d}{dt}f(t, t)\cdot \frac{\partial f}{\partial y}(f(t, -t), f(t, t))\\
=\left(\frac{dt}{dt}\cdot \frac{\partial f}{\partial x}(t, -t)+\frac{d(-t)}{dt}\cdot\frac{\partial f}{\partial y}(t, -t)\right)\cdot\frac{\partial f}{\partial x}(f(t, -t), f(t, t))\\
+\left(\frac{dt}{dt}\cdot \frac{\partial f}{\partial x}(t, t)+\frac{dt}{dt}\cdot\frac{\partial f}{\partial y}(t, t)\right)\cdot\frac{\partial f}{\partial y}(f(t, -t), f(t, t))\\
=\left(\frac{\partial f}{\partial x}(t, -t)- \frac{\partial f}{\partial y}(t, -t)\right)\cdot\frac{\partial f}{\partial x}(f(t, -t), f(t, t))
+\left(\frac{\partial f}{\partial x}(t, t)+\frac{\partial f}{\partial y}(t, t)\right)\cdot\frac{\partial f}{\partial y}(f(t, -t), f(t, t))
$$
Remember that $\frac{\partial f}{\partial x}$ specifically means "The partial derivative of $f$ with respect to the first variable", and technically has little to do with the actual variable $x$ (which doesn't even appear here). In the same way, $\frac{\partial f}{\partial y}$ means the partial derivative with respect to the second variable.
This is the real meat of the problem. For part 2. inserting $t=0$ and assuming $f(0,0)$ allows a lot of simplification and shouldn't be too problematic. And part $3$ should not be too much trouble either.
