Prove that a function of two variables is differentiable at $(0,0)$ I consider the function defined by
$$
\left\{
    \begin{array}{ll}
                \frac{x^2y^2}{x^2+y^2}\quad\text{if}\quad(x,y)\neq (0,0)\\
                0\quad\quad\;\;\;\text{if}\quad (x,y) = (0,0)
            \end{array}
            \right.
$$
I would like to show that this function is differentiable at $(0,0)$.
My attempt :
Using the fact that if a function  $f :\mathbb{R}^n\to\mathbb{R}$ is differentiable at a point $x_0$ if and only if
$$
\varepsilon(h) = \frac{1}{\lVert h\rVert}\left(f(x_0+h) -f(x_0) -df(x_0)(h) \right)\xrightarrow[h \to 0]{} 0
$$
I consider the function
$$
\varepsilon(h_1,h_2) = \frac{1}{\sqrt{h_1^2+h_2^2}}\left(\frac{h_1^2h_2^2}{h_1^2+h_2^2 }\right) = \frac{h_1^2h_2^2}{(h_1^2+h_2^2)^{3/2}}
$$
Clearly we have
$$
\lvert h_1\rvert \leq (h_1^2 + h_2^2)^{1/2}\quad\text{and}\quad h_2^2\leq h_1^2 + h_2^2\implies\lvert h_1\rvert h_2^2\leq(h_1^2 + h_2^2)^{3/2}
$$
which implies
$$
0\leq \lvert\varepsilon(h_1,h_2)\rvert = \frac{h_1^2h_2^2}{(h_1^2+h_2^2)^{3/2}}\leq\frac{h_1^2h_2^2}{\lvert h_1\rvert h_2^2} = \lvert h_1\rvert\xrightarrow[(h_1,h_2) \to (0,0)]{} 0
$$
Is this seems correct to you ?
Thank you a lot !
 A: It’s almost perfect. The only issue I have is in the last two steps, particularly when you write $\leq \frac{h_1^2h_2^2}{|h_1|h_2^2}$. This is troublesome because your original assumption is only that $(h_1,h_2)\neq (0,0)$, meaning that $h_1,h_2$ cannot be zero simultaneously. But, it is very much possible that $h_1\neq 0$ and $h_2=0$ (or vice-versa). As such, writing $h_2^2$ in the denominator is wrong (sure, here you have a $h_2^2$ in the numerator, which cancels things out, but keep in mind that you can only ‘cancel’ when you have non-zero numbers).
Here’s how I’d present the inequalities. We have for all $(h_1,h_2)\neq (0,0)$,
\begin{align}
|\epsilon(h_1,h_2)|&=\frac{h_1^2h_2^2}{(h_1^2+h_2^2)^{3/2}}=|h_1|\cdot\frac{|h_1|}{(h_1^2+h_2^2)^{1/2}}\cdot\frac{h_2^2}{h_1^2+h_2^2}\leq |h_1|\cdot 1\cdot 1=|h_1|.
\end{align}
The fact that the fractions are at most $1$ is obvious since the numerators are smaller or equal to the denominators.
Of course, you can do things slightly differently and prove that $|\epsilon(h_1,h_2)|\leq |h_2|$ if you wanted to. By presenting the inequalities this way, notice that I have only kept $h_1^2+h_2^2$ in the denominator, which by my assumption of $(h_1,h_2)\neq (0,0)$ is non-zero (in fact strictly positive), so I avoid the fatal mistake of division by zero.
