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?

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.
