1. **State the problem:** We need to find $\frac{dy}{dx}$ using implicit differentiation for the equation $$3y^2 + \tan^3 x = x^2 + 4xy.$$\n\n2. **Recall the formula and rules:** Implicit differentiation means differentiating both sides of the equation with respect to $x$, treating $y$ as a function of $x$. Use the chain rule for $y$ terms: $$\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}.$$ Also, use product rule for terms like $4xy$: $$\frac{d}{dx}[uv] = u'v + uv'.$$\n\n3. **Differentiate each term:**\n- Left side:\n - $\frac{d}{dx}[3y^2] = 3 \cdot 2y \frac{dy}{dx} = 6y \frac{dy}{dx}$\n - $\frac{d}{dx}[\tan^3 x] = 3 \tan^2 x \cdot \sec^2 x$ (chain rule for $\tan^3 x$)\n- Right side:\n - $\frac{d}{dx}[x^2] = 2x$\n - $\frac{d}{dx}[4xy] = 4 \left(1 \cdot y + x \frac{dy}{dx}\right) = 4y + 4x \frac{dy}{dx}$\n\n4. **Write the differentiated equation:**\n$$6y \frac{dy}{dx} + 3 \tan^2 x \sec^2 x = 2x + 4y + 4x \frac{dy}{dx}.$$\n\n5. **Group $\frac{dy}{dx}$ terms on one side:**\n$$6y \frac{dy}{dx} - 4x \frac{dy}{dx} = 2x + 4y - 3 \tan^2 x \sec^2 x.$$\n\n6. **Factor out $\frac{dy}{dx}$:**\n$$\left(6y - 4x\right) \frac{dy}{dx} = 2x + 4y - 3 \tan^2 x \sec^2 x.$$\n\n7. **Solve for $\frac{dy}{dx}$:**\n$$\frac{dy}{dx} = \frac{2x + 4y - 3 \tan^2 x \sec^2 x}{6y - 4x}.$$\n\n**Final answer:**\n$$\boxed{\frac{dy}{dx} = \frac{2x + 4y - 3 \tan^2 x \sec^2 x}{6y - 4x}}.$$