I want to prove that $f(x,y) = \frac{x^2y^2}{x^2+y^2} , (f(0,0) = 0)$ is (totally) differentiable at $ \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. I want to use the criterion that a function which is continuously partially differentiable is differentiable. So I compute $\frac{\partial f}{\partial x}=\frac{2xy^4}{(x^2+y^2)^2}$ (The partial derivative $\frac{\partial f}{\partial y}$ works analogously). Going back to the definition of partial deriviative with respect to the $x$ direction, we see that $\frac{\partial f}{\partial x}(0,0)=0$.
So it remains to show that $\lim_{(x,y) \to (0,0)} \frac{\partial f}{\partial x}(x,y) = 0.$ I use the $\epsilon-\delta$ criterion: Let $\epsilon >0 $. Let $\delta := \frac{1}{2} \epsilon$. So let $d((x,y),(0,0)) = \sqrt{x^2+y^2}< \delta$, say $\sqrt{x^2+y^2} = \delta'<\delta.$ This implies $|x|\leq\delta'<\delta, |y|\leq\delta'<\delta$. Therefore: $|\frac{\partial f}{\partial x}(x,y)| \leq \frac{2(\delta')^5}{(\delta')^4} = 2\delta'<2\delta=\epsilon$, as desired.
Is this analysis correct?
Your idea is fine but $0<\sqrt{x^2+y^2}<\delta$ doesn't imply $$\frac{1}{(x^2+y^2)^2}<\frac{1}{\delta^4} $$ The correct inequality is actually in the opposite direction. Furthermore, you have to show $$\left|\frac{2xy^4}{(x^2+y^2)^2}\right|<\varepsilon $$ (The absolute value is important here.) I would instead use $|xy|\leq x^2+y^2$ and $y^2\leq x^2+y^2$ so $$\left|\frac{2xy^4}{(x^2+y^2)^2}\right|\leq \frac{2|y|^3}{x^2+y^2}\leq 2|y| $$ Then, $|y|\leq \sqrt{x^2+y^2}<\varepsilon/2$, and you can finish from here. So, your choice of $\delta$ works here.