Second order ODE solution verification

Question

Knowing that $x'(-3) = 1$ and $x(-3)=0$ solve $$ 3x'x'' = e^x $$ I start with letting $u(x)=x'$. Plugging that in into the equation leaves me with: $$ u u' = \frac{e^x}{3u} $$ $$ 3u^2 du = e^x dx $$ integrate that to get $$ (x')^3 = u^3 = e^x + c $$ I suppose I should use here the fact that $x'(-3) = 1$ and $x(-3) = 0$: $$ 1^3 = e^{x(-3)} + c \\ 1 = 1 + c \implies c = 0 $$ so $$ (\frac{dx}{dt})^3 = e^x\\ \frac{dt}{dx} = e^{-\frac{x}{3}}\\ dt = e^{-\frac{x}{3}}dx\\ t = -3e^{-\frac{x}{3}} + c $$ Plug in $x(-3) = 0$ again: $$ -3 = -3e^0 + c \implies c = 0 $$ So the final solution is $$ t = -3e^{-\frac{x}{3}} $$ which can be modified to: $$ (-\frac{1}{3}t)^{-3} = e^x\\ x(t) = -3\ln(-\frac{1}{3}t) $$ Is that a correct solution?