I need to evaluate the integral \begin{equation} \int_0^1\mathrm dt\,f\left(t\right)\delta^{\left(3\right)}\left(\vec r\left(t\right)-\vec r_0\right) \end{equation} where there is only one $0\leq t_0\leq 1$ such that $\vec r\left(t_0\right)=\vec r_0$. I need to solve the integral in terms of $t_0$ and $\vec r^{\left(i\right)}\left(t_0\right)$ ($\vec r\left(t\right)$ and its derivatives at $t_0$).
I tried evaluating as a product of three delta functions, but that took me nowhere.
Forgive my previous manifestly incorrect answer; this should be correct. $$\begin{align}\int_0^1\mathrm{d}t f(t)\delta^{(3)}(\mathbf{r}(t)-\mathbf{r}_0) & =\int^1_0 \mathrm{d}t f(t)\delta(x(t)-x_0)\delta(y(t)-y_0)\delta(z(t)-z_0) \\ & =\int_0^1 \mathrm{d}tf(t)\frac{\delta(t-t_0)}{|x'(t_0)|}\frac{\delta(t-t_0)}{|y'(t_0)|}\frac{\delta(t-t_0)}{|z'(t_0)|} \\ & = \int_0^1\mathrm{d}t \frac{f(t)}{|x'(t_0)y'(t_0)z'(t_0)|}\delta(t-t_0) \\ & = \frac{f(t_0)}{|x'(t_0)y'(t_0)z'(t_0)|} \end{align}$$
Not a separate answer, but just as a basic illustration what the mapped delta function means. Since $x(t)=x_0$ only for a single $t_0$ (but you can generalize this to a number of roots), $\delta(x(t)-x_0)$ is certainly zero for almost all $t$. The only thing that interests you is the neighbourhood of $t_0$, because only there the delta function deviates from zero.
Hence you can approximate $x(t)$ by a Taylor expansion around $t_0$: $$x(t)\approx x(t_0)+x^\prime(t_0)\cdot (t-t_0)= x_0+\eta\cdot (t-t_0)$$ Then $$\delta(x(t)-x_0)=\delta(\eta\cdot (t-t_0))$$ Suppose $\eta$ was positive. By a simple linear scaling of the integration variable you can now calculate the behavior of that distribution: $$\int \phi(t)\delta(x(t)-x_0)dt=\int \phi(t)\delta(\eta\cdot (t-t_0))dt=$$ $$=\int \phi(t)\frac{\delta(\eta\cdot (t-t_0))}{\eta}\eta d(t-t_0)=\int \phi(t(\tau))\frac{\delta(\tau)}{\eta}d\tau=\frac{1}{\eta}\phi(t(\tau=0))$$ where $\tau=\eta\cdot (t-t_0)$ has been introduced. Since $\tau=0$ is equivalent to $t=t_0$ it follows that $$\int \phi(t)\delta(x(t)-x_0)dt=\frac{1}{\eta}\phi(t_0)=\int \phi(t)\frac{\delta(t-t_0)}{\eta}dt$$ for any arbitrary test function $\phi(t)$. If $\eta$ is negative, the change of variables flips direction of the integral, so if you want to keep the integration direction (which you normally do for easier notational reasons...), you have to account for the sign before the integral, i.e. $$\int \phi(t)\delta(x(t)-x_0)dt=-\frac{1}{\eta}\phi(t_0)=\int \phi(t)\frac{\delta(t-t_0)}{-\eta}dt$$ With $\eta=x^\prime(t_0)$, the relation to prove is written more compactly as: $$\delta(x(t)-x_0)=\frac{\delta(t-t_0)}{|x^\prime(t_0)|}$$
