If u solves the equation $$ u_{t} = \frac{u_{xx}}{u_{x}^2}$$
in $\mathbb{R} \times (0,\infty)$and $v$ is the inverse of $u$ in $x$, as in $y=u(x,t)$ iff $x = v(y,t)$. I need to be able to show that $v$ satisfies a linear PDE.
I've been playing around with the chain rule for a while and can't seem to get anywhere. Either that or I'm applying it wrong. Perhaps there is a different approach?
You're right to use the chain rule. The key equation to remember is $$x = v(u(x,t),t).$$ Taking $t$-partials of this equation gives $$0 = v_t(u(x,t),t) + v_y(u(x,t),t)u_t(x,t).$$ We'll hereafter suppress $t$ as a variable. Rearrange to get $$v_t(y) = -v_y(y) u_t.$$ Taking $x$-partials of the key equation yields $$1 = v_y(u(x))u_x,$$
or $v_y(y) = 1/u_x(x)$.
Substituting this into the third display yields
$$v_t(y) = -\frac{u_t}{u_x}.$$
Recall that your given PDE is the following:
$$u_t = \frac{u_{xx}}{u_x^2} = -\frac{\partial}{\partial x}\Big( \frac 1 {u_x}\Big) = -\frac{\partial}{\partial x}v_y(u(x)) = - v_{yy}(y)u_x(x).$$
Plugging this into the preceding display, finally,
$$v_t = v_{yy}.$$
