Jtdylktuy

it is known that $f(0,0) = 0$ and $f$ is differentiable at $(0,0)$. It is also known that for every $t>0$ we have $f(cos(t)/t , sin(t)/t) > 0$.

Show that necessarily $operatorname{Gradient}(0,0) = (0,0)$.

I have no clue how to solve this. I have tried to show through limits, I have tried to get something from the definition ( that with $varepsilon, :(x/y) to 0$ at the limit)

I am sorry for the bad latex language really.

Suppose that the gradient of $f$ at $(0,0)$ exists and is not $(0,0).$
That would imply that if you look at a small enough neighborhood of $(0,0),$
within that neighborhood $f$ would look like a linear function (informally speaking), and that linear function would be increasing in some directions from $(0,0)$ and decreasing in other directions from $(0,0).$

Now ask how $f$ can look like a decreasing function in some directions from $(0,0)$ and still have $fleft(frac{cos t}{t},frac{sin t}{t}right) > 0$ for every positive real number $t.$
If the answer is that it cannot, then $f$ cannot have a non-zero gradient at $(0,0).$

The problem of writing the proof is mainly how to relate the given facts to the formal definition of the gradient and the theorems about it.