While there is no trouble finding the formula for functions with distinct simple roots
$$\delta(f(x)) = \sum_{i:f(x_i)=0} \frac{\delta(x-x_i)}{|f'(x_i)|} \tag{1}$$
e.g. on Wikipedia, I cannot find anything on functions with roots of higher multiplicity. Thus my question, how to treat that?
For each root consider the local Taylor series, truncated to the first non-vanishing term:
Let $x_i$ denote the distinct roots of $f(x)$ with a multiplicity of $n_i$ respectively. Then
$$\delta(f(x)) = \sum_i \delta\Bigg(\frac{f^{n_i)}(x_i)}{n_i!}(x-x_i)^{n_i} + \mathcal O\Big((x-x_i)^{n_i+1}\Big)\Bigg).$$
Since $\delta$ vanishes almost everywhere, the higher order terms can be dropped, and thus only $\delta(ax^n)$ needs to be determined. Change of variables yields
$$\begin{align*} \delta(ax^n)\,dx &= \delta((\sqrt[n]ax)^n)\frac{d(\sqrt[n]ax)}{|\sqrt[n]a|} \\ &= \delta((\sqrt[n]ax)^n)\frac{d(\sqrt[n]ax)^n}{|\sqrt[n]a\cdot n(\sqrt[n]ax)^{n-1}|} \qquad\Bigg|\quad ax^n \to x \\ &= \delta(x)\frac{dx}{|n\sqrt[n]ax^{\frac{n-1}n}|} \\ &= \frac{\delta(x)}{|n\sqrt[n]{ax^{n-1}}|}\,dx. \end{align*}$$
Putting it all together we then get
$$\boxed{\delta(f(x)) = \sum_i\frac{\delta(x-x_i)}{\Bigg|n_i\sqrt[n_i]{\frac{f^{(n_i)}(x_i)}{n_i!}(x-x_i)^{n_i-1}}\Bigg|}}$$
For $n_i=1$ this is consistent with $(1)$, for $n_i>1$ the singularity remains and apparently can only be cancelled when the $\delta$-distribution is applied to a function with a root of at least order $1-\frac1{n_i}$.
It might be interesting to consider $n_i\to\infty$, but it's getting late...