What are the inflections points of y = e 2 x - e x ?

By the Chain Rule, the first derivative is ${\displaystyle y\prime =2e^{2x}-e^x}$ and the second derivative is ${\displaystyle y\prime \prime =4e^{2x}-e^x}$. Inflection points occur at x values where the sign of the second derivative changes (from positive to negative or negative to positive).
Setting ${\displaystyle y\prime \prime =0}$ leads to ${\displaystyle 4e^{2x}-e^x=0}$. The left-hand side of this equation can be factored as ${\displaystyle e^x(4e^x-1)=0}$. Since ${\displaystyle e^x}$ is never zero, it follows that we just need to solve ${\displaystyle 4e^x-1=0}$ to get ${\displaystyle x=\ln \left(\frac{1}{4}\right)=-\ln \left(4\right)\approx -1.386}$.
You can check that ${\displaystyle y\prime \prime =4e^{2x}-e^x}$ changes sign at this point by graphing it. Therefore, there is an inflection point at ${\displaystyle x=-\ln \left(4\right)}$. The second coordinate of this point can be found by plugging it into the original function to get ${\displaystyle e^{2\cdot \ln \left(\frac{1}{4}\right)}-e^{\ln \left(\frac{1}{4}\right)}=\frac{1}{16}-\frac{1}{4}=-\frac{3}{16}}$. The inflection point is therefore ${\displaystyle (-\ln \left(4\right),-\frac{3}{16})}$.