gradient of f at (1,1) from directional derivatives and vectors

Question

Image of the problem

Find the derivative (gradient) of $f$ at $(1,1)$ given the following directional derivatives where $u=2i+2j$ and $v=3i+j$

$D_uf(1,1)=\frac{3}{\sqrt2}$,

$D_uf(-1,1)=\frac{7}{\sqrt2}$,

$D_vf(1,1)=-\frac{1}{\sqrt10}$,

$D_vf(-1,1)=-\frac{5}{\sqrt10}$.

Answer 1

Assuming that this function has a gradient (see my comments) we can write it as $\nabla f= g(x,y)\vec{i}+ h(x,y)\vec{j}$. Now, its derivative in the direction of $\vec{u}= \vec{i}+ 2\vec{j}$ is $2g(1,1)+ 2h(1,1)= \frac{3}{\sqrt{2}}$ and its derivative in the direction of $\vec{v}= 3\vec{i}+ \vec{j}$ is $3g(1,1)+ h(1,1)= \frac{7}{\sqrt{2}}$. That gives two equations that can be solved for g(1, 1) and h(1, 1). But as I implied in my comments, the gradient is local. Knowing the value of g and h at (1, 1) does not tell us what g and h (and so f) are for other points.

Were you given other information about f, like it being linear of a conic?