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I've found the following theorem for single variable real valued functions:

The converse of the chain rule: Suppose that $f,g$ and $u$ are related so that $f(x)=g(u(x))$. If $u$ is continuous at $x_0$, $f'(x_0)$ exists and $g'(u(x_0))$ exists and is non-zero; then $u'(x_0)$ is defined and we have: $$f'(x_0)=g'(u(x_0))u'(x_0)$$

I'm interested in knowing if this "other converse" also holds:

Suppose that $f,g$ and $u$ are related so that $f(x)=g(u(x))$. If $g$ is continuous at $u(x_0)$, $f'(x_0)$ exists and $u'(x_0)$ exists and is non-zero; then $g'(u(x_0))$ is defined and we have: $$f'(x_0)=g'(u(x_0))u'(x_0)$$

If anyone is interested here is the proof of the first one (page 12), I tryed to proved the second one in an analogous way but I didn't succeed. For my immediate purposes I would be happy with knowing it holds, and in case it doesn't some counterexample could be instructive.
I'm also interested in knowing if those statements generalises in the obvious way to functions from $\mathbb{R}^n$ to $\mathbb{R}^m$.

Thanks in advance

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    If $u:\mathbf{R}^{n} \to \mathbf{R}^{n}$, if $Du$ is _continuous_, and if $\det Du(x_{0}) \neq 0$, then $u$ is invertible in some neighborhood of $x_{0}$. If $v$ denotes an inverse of $u$, then $$ g(y) = f(v(y)) $$ for all $y$ in some neighborhood of $y_{0} = u(x_{0})$. The ordinary chain rule implies $g$ is differentiable at $y_{0}$, and that $$ Dg(y_{0}) = Df(x_{0}) Dv(y_{0}) = Df(x_{0}) Du(x_{0})^{-1}, $$ or $Df(x_{0}) = Dg(u(x_{0})) Du(x_{0})$. I don't see offhand how to handle the situation where $Du$ is discontinuous at $x_{0}$. – 2017-01-16
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    @AndrewD.Hwang what about $u:\mathbb{R}^n \to \mathbb{R}^m$ and $g:\mathbb{R}^m \to \mathbb{R}^p$? I guess in this case we can't use the Inverse Function Theorem – 2017-01-16
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    I have no idea whether the second converse holds, however, the first converse can be generalized in the obvious way. For a proof, see these notes by Bruce Driver (pages 2, 3): http://www.math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/chap22.pdf (though $X$ and $Y$ are Banach spaces, you may just ignore this fact because nowhere does this matter anyway). – 2017-01-17

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