Answer
$y=\dfrac{x+2}{3}.$
Work Step by Step
$y^2x^2+y^4=2x^2\rightarrow$
$\dfrac{d}{dx}(y^2x^2)+\dfrac{d}{dx}(y^4)=\dfrac{d}{dx}(2x^2)\rightarrow$
$(2y^2x+\dfrac{dy}{dx}(2x^2y))+\dfrac{dy}{dx}(4y^3)=4x\rightarrow$
$\dfrac{dy}{dx}=\dfrac{4x-2y^2x}{4y^3+2x^2y}=\dfrac{2x-y^2x}{2y^3+x^2y}.$
At $(1, 1)\rightarrow\dfrac{dy}{dx}=\dfrac{2(1)-(1^2)(1)}{2(1^3)+(1^2)(1)}=\dfrac{1}{3}.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-1)=\frac{1}{3}(x-1)\rightarrow y=\dfrac{x+2}{3}.$
