Inverse of a exponential function

Hint. Let $z=e^{-x/4}$ and solve the quadratic equation with respect to $z$. $$y = { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}={ 1 \over 3} z^2 + { 2 \over 3} z\Leftrightarrow (z+1)^2=1+3y.$$

P.S. As pointed out by Yves Daoust, starting from the hint we easily find $$x=f^{-1}(y)=-4\log\left(\sqrt{3y+1}-1\right)$$ which can be written also in the following way \begin{align} x=f^{-1}(y)&=-4\log\left(\sqrt{3y+1}-1\right)=4\log\left(\frac{1}{\sqrt{3y+1}-1}\right)\\ &=4\log\left(\frac{\sqrt{3y+1}+1}{(3y+1)-1}\right) =4\log\left(\frac{\sqrt{3y+1}+1}{y}\right)-4\log(3). \end{align} So there is a missing minus sign in your formula. Note that $f(0)=1$ and therefore $f^{-1}(1)=0$.

Let $z:=e^{-x/4}$. We can write the equation as

$$3y=z^2+2z.$$

The solution of this quadratic equation is

$$z=\pm\sqrt{3y+1}-1$$

or

$$x=-4\log\left(\sqrt{3y+1}-1\right)$$ because the negative solution must be rejected.