If $x^2+xy=10$, then when $x=3$, $\dfrac{dy}{dx}=$

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

$$x^2+xy=10$$ $$\frac{d}{dx}(x^2+xy)=\frac{d}{dx}(10)$$ $$\underbrace{\frac{d}{dx}(x^2)}_\text{power rule}+\underbrace{\frac{d}{dx}(xy)}_\text{product rule}=\underbrace{\frac{d}{dx}(10)}_\text{constant rule}$$ $$2x + x\cdot\frac{dy}{dx}+y\cdot 1 = 0$$

Solving for $\dfrac{dy}{dx}$,

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

To obtain the $y$ value, we can substitute $x=3$ into the initial equation.

$$x^2+xy=10$$ $$(3)^2+(3)y=10$$ $$9+3y=10$$ $$3y=1$$ $$y=\frac{1}{3}$$

Substituting $x=3$ and $y=\dfrac{1}{3}$,

$$\frac{dy}{dx}=-\frac{2x+y}{x}$$ $$=-\frac{2(3)+\frac{1}{3}}{3} =-\frac{6+\frac{1}{3}}{3} = -\frac{\frac{19}{3}}{3}$$ $$=\boxed{-\frac{19}{9}}$$

Though the question is framed in the form of implicit differentiation, we can easily solve for $y$ in terms of $x$.

$$x^2+xy=10$$ $$xy=10-x^2$$ $$y=\frac{10-x^2}{x}$$

We can break up the fraction and use the power rule to differentiate.

$$y=\frac{10}{x}-\frac{x^2}{x} = \frac{10}{x}-x$$ $$y'= -\frac{10}{x^2}-1$$ $$y' |_{x=3} = -\frac{10}{(3)^2}-1$$ $$= -\frac{10}{9}-1$$ $$= \boxed{-\frac{19}{9}}$$