1. State the problem: Simplify or analyze the equation $$x^{2} y^{2} + 3y = 4x$$ or express it in a more useful form. 2. The equation is $$x^{2} y^{2} + 3y = 4x$$. 3. We can try to solve for one variable in terms of the other. Let's isolate terms involving $y$: $$x^{2} y^{2} + 3y = 4x$$ 4. Treat this as a quadratic in $y$: $$x^{2} y^{2} + 3y - 4x = 0$$ 5. Use the quadratic formula for $y$ where $a = x^{2}$, $b = 3$, and $c = -4x$: $$y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-3 \pm \sqrt{9 + 16x^{3}}}{2x^{2}}$$ 6. The solution for $y$ in terms of $x$ is therefore: $$y = \frac{-3 \pm \sqrt{9 + 16x^{3}}}{2x^{2}}$$ This describes $y$ implicitly as functions of $x$ for values where the discriminant $$9 + 16x^{3} \geq 0$$. Final answer: $$y = \frac{-3 \pm \sqrt{9 + 16x^{3}}}{2x^{2}}$$