How are you going to solve this?

Differentiate to find $2xf(x)=f'(x)$ . Multiply by $e^{-x^2}$ to get $e^{-x^2}f'(x)-2xe^{-x^2}f(x)$ $=\frac{d}{dx}(e^{-x^2}f(x))$ $=0$ and $f(x)=Ce^{x^2}$ . Now $f(1)=e$ gives $C=1$ so $f(2)=e^4$ and $\ln(f(4))=\boxed{4}$ .

The equation reduces to a linear differential equation:

$y' = 2xy$

Solving the DE gives us $y=Ce^{x^{2}}$ Substituting our boundary conditions gives us

$y = e^{x^{2}}$

It's not really to determine $ln(f(2))$ which will give us $4$