Differential Equation $xf''(x) + 3x(f'(x))^2 = 1 - e^{-x}$

Question

Given a function $f$ which satisfies the differential equation $xf''(x) + 3x(f'(x))^2 = 1 - e^{-x}$

if $f(0) = f'(0) = 0$, find the smallest constant $A$ such that $f(x) \le Ax^2$ for all $x \ge 0$