If $x^3+3xy+2y^3=17$, then in terms of $x$ and $y$, $\dfrac{dy}{dx}=$

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Differentiate both sides of the equation with respect to $x$.

$$x^3+3xy+2y^3=17$$ $$\dfrac{d}{dx}(x^3+3xy+2y^3)=\frac{d}{dx}(17)$$

Using the appropriate rules for each term,

$$\underbrace{\dfrac{d}{dx}(x^3)}_\text{power rule}+3\underbrace{\dfrac{d}{dx}(xy)}_\text{product/chain rule}+ 2\underbrace{\dfrac{d}{dx}(y^3)}_\text{power/chain rule}=\underbrace{\frac{d}{dx}(17)}_\text{constant rule}$$ $$3x^2+3\left(x\cdot\frac{dy}{dx}+y\cdot1\right)+2\cdot3y^2\cdot\frac{dy}{dx}=0$$

We can divide all terms by 3, then solve for $\dfrac{dy}{dx}$:

$$x^2+x\frac{dy}{dx}+y+2y^2\frac{dy}{dx}=0$$ $$x\frac{dy}{dx}+2y^2\frac{dy}{dx}=-x^2-y$$ $$(x+2y^2)\frac{dy}{dx}=-(x^2+y)$$ $$\frac{dy}{dx}=-\frac{x^2+y}{x+2y^2}$$