How to find critical points of two equations? So i have this function 
$$f(x, y) = (x − y)(1 − xy)$$
which i re-write as $x-x^2y-y+xy^2$
I find the partial derivatives:
$$f_x(x,y)=1-2xy+y^2$$
$$f_y(x,y)=-x^2-1+2xy$$
I want to look for points $(a,b)$ such that $f_x(a,b)=0$ and $f_y(a,b)=0$
So i will write $x$ in terms of $a$ and $y$ in terms of $b$ .
So i write:
$$1-2ab+b^2=0$$
and 
$$-a^2-1+2ab=0.$$
I have dyscalculia and have no clue how to proceed from here...the values i get are weird and do not make sense...
I read somewhere that when you are not able to take the critical points of an equation , then you might want to right it in another form ? Is that possible?
Or if i'm saying nonesense, can somebody please help me and show me some simple steps on how to get to the critical points, with explanations on why do you such steps? On the  answer sheet , the critical points are:
$P_1(1, 1)$ and $P_2(−1,−1)$
Thanks!!
 A: From 
\begin{align}
1-2ab+b^2&=0\\
-a^2-1+2ab&=0
\end{align}
Multiple the second by $-1$ to get
\begin{align}
1-2ab+b^2&=0\\
a^2+1-2ab&=0
\end{align}
Now notice that $1 - 2ab$ appears in both, so write
\begin{align}
1-2ab&= -b^2\\
1-2ab&=-a^2
\end{align}
from which you can conclude that $a^2 = b^2$, which gives you two cases: 


*

*$a = b$

*$a = -b$.


For the first, your first equation becomes
\begin{align}
1-2ab+b^2&=0\\
1-2a^2+a^2&=0\\
1-a^2&=0\\
a^2&=1\\
a&=\pm 1
\end{align}
and hence leads to the solutions $(1,1)$ and $(-1, -1)$. 
Now YOU try to do the same thing for the second case, substituting $-a$ for $b$ in the equations to see what you get. 
A: Hint: Write your system in the form
$$1+y^2=2xy$$
$$1+x^2=2xy$$ so we get $$x^2=y^2$$ or $$(x-y)(x+y)=0$$
A: It appears that the equations
$1 - 2ab + b^2 = 0 \tag 1$
and
$-a^2 - 1 + 2ab \tag 2$
are correct.  We add them to obtain
$-a^2 + b^2 = 0, \tag 3$
which implies
$b = \pm a; \tag 4$
we may substitute this into (2) and find
$-a^2 -1 \pm 2a^2 = 0, \tag 5$
or
$a^2 - 1 = 0, \tag 6$
and
$3a^2 + 1 = 0; \tag 7$
we discard (7) since it has no real solutions.  Thus
$a = \pm 1, \tag 8$
whence also
$b = \pm 1; \tag 9$
we further narrow the possibilities by checking (8), (9) against (1), and discover that the solutions are in fact $(1, 1)$ and $(-1, -1)$, which also satisfy (2).
