1. The problem is to solve the equation $ye^{-3x} - 3x = y^2$ for $y$ in terms of $x$. 2. This is a nonlinear equation involving both $y$ and $x$. We want to isolate $y$ if possible. 3. Rewrite the equation: $$ye^{-3x} - y^2 = 3x$$ 4. Rearrange to standard quadratic form in $y$: $$y^2 - ye^{-3x} + 3x = 0$$ 5. Use the quadratic formula for $y$: $$y = \frac{e^{-3x} \pm \sqrt{(e^{-3x})^2 - 4 \cdot 1 \cdot 3x}}{2} = \frac{e^{-3x} \pm \sqrt{e^{-6x} - 12x}}{2}$$ 6. The discriminant $e^{-6x} - 12x$ must be non-negative for real solutions. 7. Therefore, the solutions for $y$ are: $$y = \frac{e^{-3x} \pm \sqrt{e^{-6x} - 12x}}{2}$$ This completes solving for $y$ in terms of $x$.