Why is the possibility $x-y = 0$ rejected? I'll state the question from my textbook below:

If $x (1+y)^{1/2} + y(1+x)^{1/2} = 0$, for, $-1<x<1$, prove that
$\frac{dy}{dx} = - \frac{1}{(1+x)^2}$

Here's how I tried proving it:
$x (1+y)^{1/2} + y(1+x)^{1/2} = 0$
Rearranging and squaring we get:
$x^2(1+y) = y^2(1+x)$
Simplifying and factorizing we get:
$(x+y+xy)(x-y) = 0$
Case I:
$x+y+xy = 0$
From here we get the desired equation. Since the proof is irrelevant to my question, I'll skip it. But I note that if I replace the value of $y$ I get from this equation is in harmony with the original equation given in the question.
Case II:
$x-y = 0$
Clearly in this case,
$\frac{dy}{dx} = 1$
This is where the problem arises. Isn't this a valid answer too? Probably not, because the value of $y$ I get from this equation satisfies the original equation only for $x = -1,0$. Why is this? Here are the questions I'm looking answer for:
1) Is something wrong with squaring the equation?
2) Is there a mathematical reason to reject $x-y = 0$ as a possibility?
I am also interested in alternative methods to prove the required equation. So if you have any, feel free to post them as an answer. Any help would be appreciated.
 A: On putting $y=x$, left side of equation is always positive or negative. Its satisfied only for $(0,0)$. While squaring, there may be some extra conditions introduced, which may not be satisfied by original equation.
As an example, consider the simple straight line $y=x$, squaring will give $y^2 = x^2$ or $y=\pm x$ which is a pair of straight lines.
On the other hand, if you take all things to one side so that there is $0$ on the other side, squaring is justified. That will not introduce any new roots. $y-x = 0 \iff (y-x)^2 = 0$
A: We reject $x-y=0$ because then $y = x$. Now we plug this back into the original equation and find it is not a solution because, as you've stated, we don't get a true statement for all x in (-1, 1).  That is the part that is being overlooked by you.  The potential solution arising from $x -y =0$ does not satisfy the hypotheses of the original problem and therefore is not a solution.
A: Let try to plug in $x=y$ in the original equation to verify that that solution doesn't work. 
This means that by squaring we have obtained an extra solution to discard.
