Solving the equation $- y^2 - x^2 - xy = 0$ Ok, this is really easy and it's getting me crazy because I forgot nearly everything I knew about maths!
I've been trying to solve this equation and I can't seem to find a way out.
I need to find out when the following equation is valid:
$$\frac{1}{x} - \frac{1}{y} = \frac{1}{x-y}$$
Well, $x \not= 0$, $y \not= 0$, and $x \not= y$ but that's not enough I suppose.
The first thing I did was passing everything to the left side:
$$\frac{x-y}{x} - \frac{x-y}{y} - 1 = 0$$
Removing the fraction:
$$xy - y² - x² + xy - xy = 0xy$$
But then I get stuck..
$$- y² - x² + xy = 0$$
How can I know when the above function is valid?
 A: Your work goes a long way towards the answer. I will assume that you are looking  for solutions in real numbers $x$ and $y$.
You want to solve $x^2-xy+y^2=0$. Note that
$$x^2-xy+y^2=\left(x-\frac{y}{2}\right)^2 +\frac{3}{4}y^2.\qquad(\ast)$$
The above result is easy to verify by expanding the right-hand side.  But it was not obtained by magic: It is a standard application of the powerful idea usually called Completing the Square.  You have undoubtedly met this idea  earlier in other contexts. 
On the right-hand side of $(\ast)$ we have a square, namely $\left(x-\frac{y}{2}\right)^2$,   plus $3/4$ of $y^2$.  The square of the real number $y$ is always $\ge 0$.  So the only way we can satisfy the equation $x^2-xy+y^2=0$ is by taking $y=0$ and $x-y/2=0$, meaning that $x=0$.  These values are, as you pointed out, forbidden, so the original equation has no real solutions.
Remark: Through unhappy experiences, I have somewhat of an aversion to fractions, so would prefer to say that the equation $x^2-xy+y^2=0$ is equivalent to $4x^2-4xy+4y^2=0$. But 
$$4x^2-4xy+4y^2=(2x-y)^2+3y^2.$$
Or else, if we feel bad about breaking symmetry, we can avoid completing the square, and instead note that
$$2x^2-2xy+2y^2=(x-y)^2+x^2+y^2.$$
Again, we have a sum of squares on the right, and this can be $0$ only if $x$, $y$ (and therefore $x-y$) are all $0$.
Much more mechanically, we can use the Quadratic Formula. For any fixed $y$, the solutions of $x^2-xy+y^2=0$ are 
$$x=\frac{y \pm\sqrt{-3y^2}}{2}.$$
If $y\ne 0$, the number under the square root sign is negative, and therefore $\sqrt{-3y^2}$ is not a real number, so $x$ is not a real number.
A: $y=x\left(\dfrac{1 \pm\sqrt{-3}}{2}\right)$, combined with $0 \not = y\not =x\not = 0$, answers the question "when the following equation is valid".  The first statement is equivalent to $x=y\left(\dfrac{1 \mp \sqrt{-3}}{2}\right)$.
It is valid for all pairs of complex numbers with this property; it is not valid for any pair of real numbers.  
A: $x^2-xy+y^2=(x+jy)(x+j^2y)$ so $x=y(1+\sqrt{-3})/2$
