## Calculus 10th Edition

$\dfrac{dy}{dx}=0.$
$\dfrac{d}{dx}((x^2+y^2)^2)=\dfrac{d}{dx}(4x^2y)\rightarrow$ Using the Chain Rule with $u=x^2+y^2\rightarrow\dfrac{du}{dx}=2x+\dfrac{dy}{dx}(2y)\rightarrow$ $\dfrac{d}{dx}((x^2+y^2)^2)=2(x^2+y^2)(2x+\dfrac{dy}{dx}(2y)).$ $2(x^2+y^2)(2x+\dfrac{dy}{dx}(2y))=\dfrac{dy}{dx}(4x^2)+8xy\rightarrow$ $\dfrac{dy}{dx}(x^2-x^2y-y^3)=xy^2-2xy+x^3\rightarrow$ $\dfrac{dy}{dx}=\dfrac{xy^2-2xy+x^3}{x^2-x^2y-y^3}.$ At $(1, 1)\rightarrow\dfrac{dy}{dx}=\dfrac{(1)(1^2)-2(1)(1)+(1)^3}{1^2-(1^2)(1)-1^3}=0.$