How do I calculate $\dfrac{\partial}{\partial t} \left(\dfrac{\partial f(x_1(t), x_2(t))}{\partial x_1}\right)$?
Assume the necessary differentiability conditions.
I think it's $\dfrac{\partial^2 f}{\partial x_1 \partial x_2} \dfrac{\partial x_2}{\partial t} + \dfrac{\partial^2 f}{\partial x_1^2} \dfrac{\partial x_1}{\partial t}$, but I am not sure. Is that correct?
As an application of the chain rule, we have for $f:\mathbb R^n\to \mathbb R^m$ which is of class $C^1$ the following: $$\frac{d}{dt} f(x_1(t),\dots,x_n(t))= \sum_{i=1}^n \partial_i f(x_1(t),\dots,x_n(t))\frac{dx_i}{dt}(t). $$ Where $\partial_j$ is the partial derivative w.r.t. the $j$-$th$ variable. Note the multiplication here is well defined since $\frac{dx_i}{dt}(t)$ is just a scalar.
So yes, your answer is right if $f$ is of class $C^2$, why? If $f$ is $C^2$, then $$\frac{\partial^2 f}{\partial x_1 \partial x_2} = \frac{\partial^2f}{\partial x_2 \partial x_1}.$$ Your mistake could be: $$\frac{\partial }{\partial x_2} \Big(\frac{\partial f}{\partial x_1}(x_1(t),x_2(t))\Big) = \frac{\partial^2f}{\partial x_1\partial x_2}(x_1(t),x_2(t)),$$ if $f$ is not of class $C^2$.