Implicit differentiation of $(x^2 + y^2)^2 = (x-y)^2$ $$
(x^2 + y^2)^2 = (x-y)^2$$
Wolfram alpha yields this answer:
$$
y'(x) = \frac{(-2 x^3-2 x y^2+x-y)}{((2 x^2-1) y+x+2 y^3)}$$
But it's impossible to get $-y$ in the denominator
Actually, my answer is pretty much the same, except it's $+ y$ in the denominator
When differentiating the left side, you'll get:
$$(2x^2 + 2y^2) * (2x + 2yy') = 4x^3 + 4x^2 y y' + 4y^2x+4y^3 y'$$
dividing by 2 both sides and taking y' out, the left side would look:
$$y'(2x^2 y + 2y^3 + x + y)$$
as you can see, it's $+y$
so why does wolfram alpha says otherwise?
 A: $$(x^2+y^2)^2=(x-y)^2$$
$$x^4+2x^2y^2+y^4=x^2-2xy+y^2$$
$$4x^3+4xy^2+4x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}=2x-2y-2x\frac{dy}{dx}+2y\frac{dy}{dx}$$
$$y'(4x^2y+4y^3)+4x^3+4xy^2=y'(-2x+2y)+2x-2y$$
$$y'(4x^2y+4y^3+2x-2y)=-4x^3-4xy^2+2x-2y$$
$$y'=\frac{-4x^3-4xy^2+2x-2y}{4x^2y+4y^3+2x-2y}$$
Take out factor of $2$
$$y'=\frac{-2x^3-2xy^2+x-y}{2x^2y+2y^3+x-y}$$
$\large\color{green}{\checkmark}$Verified by Wolframalpha 
A: You start off with
$$
(x^2+y^2)^2=(x-y)^2 \Longleftrightarrow x^4+2x^2y^2+y^4=x^2-2xy+y^2.
$$
Now pull everything with a $y$ in it to the LHS and terms only with $x$ to the RHS:
$$
y^4-y^2+2x^2y^2+2xy=x^2-x^4.
$$
Now implicitly differentiation everything (using the product rule twice):
$$
4y^3y'-2yy'+(4xy^2+4x^2yy')+(2y+2xy')=2x-4x^3.
$$
Now factor out the $y'$:
$$
y'(4y^3-2y+4x^2+2x) = 2x-4x^3-2y-4xy^2.
$$
Isolate $y'$ (and divide everything by $2$ once you have done this), and you will see that
$$
y'=\frac{x-2x^3-2xy^2-y}{2y^3-y+2x^2y+x},
$$
and this matches the output produced by Wolfram|Alpha.
