Given a differentiable function $f$, some $x_j$ ($j \in \{1, ..., n\}$) such that $f(x_j) = 0$ $\forall j$ and $f'(x_j) \ne 0$ $\forall j$, and the following definition of the composition of a distribution $T$ with a function $f$:
$$(T \circ f)[\phi] := T\left(\left(\phi\ \circ\ f^{-1}\right)\left|\frac{\partial f^{-1}}{\partial x}\right|\right)$$
I need to prove that $$\delta\ \circ\ f = \sum_{j=1}^{n}{\frac{\delta_{x_j}}{\left|f'(x_j)\right|}}$$
My first thought was that if $f$ has multiple roots $x_i$ and $x_j$ with $x_i \ne x_j$, then it cannot be bijective, so $f^{-1}$ does not exist.
I considered defining some $\delta_n$ with $\delta_n \to \delta$ for $n \to \infty$ so that I could consider the limit of integral $\int_{-\infty}^{\infty} \delta_n(f(x))\phi(x)\ dx$. My idea was to split it up into multiple parts like this:
$$ \int_{-\infty}^{\infty} \delta_n(f(x))\phi(x)\ dx = \int_{-\infty}^{x_1 - \epsilon} \delta_n(f(x))\phi(x)\ dx + \sum_{j=1}^{n-1} \left(\int_{x_j - \epsilon}^{x_j + \epsilon} \delta_n(f(x))\phi(x)\ dx + \int_{x_j + \epsilon}^{x_{j+1} - \epsilon} \delta_n(f(x))\phi(x)\ dx \right) + \int_{x_n-\epsilon}^{x_n+\epsilon} \delta_n(f(x))\phi(x)\ dx + \int_{x_n+\epsilon}^{\infty} \delta_n(f(x))\phi(x)\ dx $$
By substituting $u = f(x)$ we have $dx = \frac{du}{f'(x)} = \frac{du}{f'(f^{-1}(u))}$, and the integrals would look like $\int_{x_j-\epsilon}^{x_j+\epsilon} \delta_n(u)\phi(f^{-1}(u)) \frac{1}{f'(f^{-1}(u))} \ du $.
Since $f^{-1}(u) = x_j$, by taking the limit $n \to \infty$ the integral is equal to $\frac{\phi(x_j)}{f'(x_j)}$, which is $\frac{\delta_{x_j}}{f'(x_j)}$.
In the other integrals, $x$ is never $x_j$, so when $n \to \infty$, the integral is always $0$, so in the end we only have the sum $\sum_{j=1}^{n}{\frac{\delta_{x_j}}{\left|f'(x_j)\right|}}$.
My problem now is that I cannot prove the existence of $f^{-1}$ in the interval $[x_j - \epsilon, x_j + \epsilon]$. If $f$ were continuously differentiable, I could choose $\epsilon$ small enough so that $f'(x_j) \ne 0$ in the whole interval, so $f'$ is strictly monotonically increasing (or decreasing) on the interval, so $f^{-1}$ exists. However, it is only given that f is differentiable.
Another problem is that in my "proof", I don't take the absolute value of $f'(x_j)$. I feel like this is because I didn't exactly use the provided definition of $T\ \circ\ f$, but I didn't have another idea.
Is there a way to "save" this proof?