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