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Differentiate both sides of the equation $3x^2y + xy^2 = 5$ with respect to $x$. Remember to use the product rule and the chain rule for terms involving $y$.$\frac{d}{dx}(3x^2y) + \frac{d}{dx}(xy^2) = \frac{d}{dx}(5)$Using the product rule for $3x^2y$: $(6x)y + 3x^2(\frac{dy}{dx})$Using the product rule for $xy^2$: $(1)y^2 + x(2y\frac{dy}{dx})$The derivative of the constant $5$ is $0$.So, we have: $6xy + 3x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$Group terms with $\frac{dy}{dx}$ on one side and other terms on the other side:$3x^2\frac{dy}{dx} + 2xy\frac{dy}{dx} = -6xy - y^2$Factor out $\frac{dy}{dx}$:$\frac{dy}{dx}(3x^2 + 2xy) = -(y^2 + 6xy)$Solve for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{y^2 + 6xy}{3x^2 + 2xy}$.