<s>[INST] total differentiability of $\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$ if the denominator equals $0$ Let $f:\mathbb{R}^3\to\mathbb{R}$ be given by $f(x,y,z)=\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$ if the denominator is not equal to $0$ and otherwise by $f(x,y,z)=0$.
Find all the points where $f$ is differentiable.

$f$ is surely differentiable when the denominator is not equal to zero so we only need to consider the case that $(x^2+y^2)(y^2+z^2)=0$.
However, even if we can show that the partial derivatives of $f$ are not continuous at those points, that still doesn't imply that $f$ is not differentiable there. How do I continue from here?
 [/INST] Choose $(x,y,z) \neq 0$ and look at $f((tx,ty,tz)) = t^3\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$. From this, if $f$ is differentiable, we would expect that
$Df(0) = 0$.
Note that ${x^2 \over x^2+y^2} \le 1$ and ${z^2 \over z^2+y^2} \le 1$ so
$|f((x,y,z))| \le |xz^2|$.
Note that $|xz| \le {1 \over 2} (x^2+z^2)$.
Suppose $|z| \le 2$, then we have $|f((x,y,z))| \le x^2+z^2 \le x^2+y^2+z^2= \|(x,y,z)\|^2$.
Hence $f$ is (Frechet) differentiable at $0$ with derivative $0$.
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