The following is a proof from my textbook:
If $u(x, t)$ is a solution of $(1)$, so is the dilated function $u(\sqrt{a}x, at)$, for any $a > 0$. Prove this by the chain rule: Let $v(x, t) = u(\sqrt{a} x, at)$. Then $v_{t} = [\partial (at)/\partial t]u_{t} = au_{t}$ and $v_{x} =[\partial(\sqrt{a} x)/\partial x]u_{x} = \sqrt{a} u_{x}$ and $v_{xx} = \sqrt{a}\cdot \sqrt{a}u_{xx} = a u_{xx}$.
I was wondering how we get that $v_{t} = [\partial (at)/\partial t]u_{t} = au_{t}$, for example? Since writing out the right-hand side of $v_{t}$ we have $\dfrac{\partial (at)}{\partial t}\cdot \dfrac{\partial u}{\partial t}$, but $\partial u$ and $\partial t$ don't "cancel out."
Hint: Write $v=u\circ g$ where $g(x,t)=(\sqrt{a} x, at)$. Then, $\frac{\partial v}{\partial t}$ can be computed by reading off from the product of the Jacobians $\mathcal J(u)\circ \mathcal J(g)=\mathcal J(v)=\begin{pmatrix} \frac{\partial v}{\partial x} & \frac{\partial v}{\partial t} \end{pmatrix}=\begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial t} \end{pmatrix}\begin{pmatrix} \frac{\partial g_1}{\partial x} & \frac{\partial g_1}{\partial t}\\ \frac{\partial g_2}{\partial x} & \frac{\partial g_2}{\partial t} \end{pmatrix}$.