Answer
$\dfrac{dy}{dx}=-\dfrac{(x+y) }{(x+2y^3)}$
Work Step by Step
Apply Implicit Differentiation Theorem: It states that for a function $F$ to be differentiable on its domain when it satisfies the following condition such that $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ and $F(x,y)=0$ defines $y$ as a differentiable function of $x$ and also, $F_y \ne 0$.
We are given that $F(x,y)=\sqrt {x^2+2xy+y^4}-3$
Now, we have: $\dfrac{dy}{dx}=-\dfrac{\frac{2x+2y}{2\sqrt {x^2+2xy+y^4}}}{\frac{2x+4y^3}{2\sqrt {x^2+2xy+y^4}}}$
or, $\dfrac{dy}{dx}=-\dfrac{(x+y) \sqrt {x^2+2xy+y^4}}{(x+2y^3)\sqrt {x^2+2xy+y^4}}$
or, $\dfrac{dy}{dx}=-\dfrac{(x+y) }{(x+2y^3)}$
