Find the value(s) of dy/dx for x^2 * y + y^2 = 5 at y =1.
What are you having trouble with? Differentiate both sides with respect to x and solve for $\displaystyle y'$. Find the cooresponding x coordinate (since you have y = 1) and plug it in to your expression for $\displaystyle y'$ (it'll contain x's and y's).
$\displaystyle x^2y+y^2~=~5$
Tackle this in parts;
Firstly use the product rule.
Let $\displaystyle u=x^2~,~v=y$
$\displaystyle \frac{du}{dx}~=~2x~,~\frac{dv}{dx}~=~\frac{dy}{dx}$
$\displaystyle \frac{d}{dx}(x^2y)~=~x^2\frac{dy}{dx}+2xy$
$\displaystyle \implies~\frac{d}{dx}(x^2y+y^2~=~5)~\implies~x^2\f rac{dy}{dx}+2xy+2y\frac{dy}{dx}~=~0$
$\displaystyle \implies~\frac{dy}{dx}(x^2+2y)~=~-2xy$
$\displaystyle
\implies~\frac{dy}{dx}~=~\frac{-2xy}{x^2+2y}
$
Then substitute your y-value and x-value (which you can calculate using the given formula)
Just like o_O said