<s>[INST] Taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$ I am a bit confused about taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$.
$$\frac{d(x^2 + y^2)^3}{dx} = \frac{d(5x^2 y^2)}{dx} $$
Edit: Incorrect step
$$= 3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 10xy^2 + 5x^22y\frac{dy}{dx}$$
$$=  3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 5 \left(2xy^2 + x^22y\frac{dy}{dx}\right)$$
...
So I don't get how we got $2y\frac{dy}{dx}$ from $y^2$. Why it is not $2\frac{dy}{dx}$???
 [/INST] An expression $f(x,y)=g(x,y)$ is implicitly differentiated by
$$ \frac{\partial f}{\partial y} y' + \frac{\partial f}{\partial x} x' =\frac{\partial g}{\partial y} y' + \frac{\partial g}{\partial x} x'$$
but with $x'=1$ in your case.
What I get is
$$ \left(6 y (x^2+y^2)^2 \right) y' + \left( 6 x (x^2 + y^2)^2 \right) 1 = \left( 10 x^2 y \right) y' + \left(10 x y^2 \right) 1$$
which needs some simplifications.
The specific question as to why $\frac{{\rm d} }{{\rm d} x}y^2 = 2 y y'$ use the rule above to get
$$ \frac{\partial y^2}{\partial y} y' + \frac{\partial y^2}{\partial x} 1 = (2 y) y' + (0) 1 = 2 y y' $$
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