The directional derivative in the direction $(h_x,h_y)$ (with $h_y \neq 0$) at $(0,0)$ is given
by $df((0,0),(h_x,h_y)) = \lim_{t \to 0} { f(th_x,t h_y) -f(0,0) \over t} = \lim_{t \to 0} { t^3 h_x^2h_y \over t^4 h_x^4 + t^2 h_y^2}$.
It is straightforward to see that the limit is zero in all cases (look at $h_y=0$ and $h_y \neq 0$ separately).
Old incorrect answer: (Thanks to @user61681 for catching that.)
The directional derivative in the direction $(h_x,h_y)$ (with $h_y \neq 0$) at $(0,0)$ is given
by $df((0,0),(h_x,h_y)) = \lim_{t \to 0} { f(th_x,t h_y) -f(0,0) \over t} = \lim_{t \to 0} { t^2 h_x^2h_y \over t^4 h_x + t^2 h_y^2} = {h_x^2 \over h_y}$.
If $(h_x,h_y)$ is in the direction given, we have
$(h_x,h_y) = \lambda ({\sqrt{3} \over 2},{1 \over 2})$ for some $\lambda>0$, and
so $df((0,0),\lambda ({\sqrt{3} \over 2},{1 \over 2})) = {3 \over 2} \lambda$.
Your proof is correct but is, in my opinion, missing a few words here and there explaining why your procedure is correct.
As an example : it seems to me that you're implicitly using the property that the directional derivative $\partial_v f$ at a point $a$ is the dot product $\nabla f (a) \cdot v$. This is not always the case ! But it is true when the function $f$ is differentiable at $a.$ So your proof should at least contain the words " Since $f$ is differentiable at ..."
Similarly you could be elaborate a bit more on certain parts:
The minimum directional derivative is:
$$−|\nabla f(2,1)|=−|⟨1,−1⟩|=−\sqrt 2$$
Why is this true ? You don't need the write a huge paragraph of course ;) but a few extra words here and there would make your proof far more readable.
Best Answer
You can just use the definition $$ \partial_{(4/5,3/5)}f(0,0) = \lim_{t \to 0} \dfrac{f(0+4t/5, 0 + 3t/5)-f(0,0)}{t} = \frac{13}{5} $$