Change of variables in a double integral-proving injectivity I want to make the following change:
$$u=x^2 - y^2 \quad v=xy$$ where my region is in the first quadrant bounded by:
$x^2-y^2 = 3 , \quad xy=1 , \quad x^2-xy-y^2 = 1 $  . 
How can I prove this change of variables is legitimate ? (i.e., how can I prove this change is injective [one-to-one]) 
I have tried solving $$ x_1 ^2 - y_1 ^2 = x_2 ^2 -y_2 ^2 , \, x_1 y_1 = x_2 y_2 $$ but this implies $x_1 = x_2 , y_1 = y_2 $ as required, but also: $x_2 = y_1 $ for example. 
Will you please help me figure out how to prove the injectivity of this ? 
Thanks a lot
 A: Original strategy of mine
Claim
$$\left\{\begin{array}{c}
u=x^2-y^2 \\
v=xy
\end{array}\right.
\iff
\left\{\begin{array}{c}
x=\sqrt{\frac{u+\sqrt{u^2+4v^2}}{2}} \\
y=\sqrt{\frac{\sqrt{u^2+4v^2}-u}{2}}
\end{array}\right.
$$
Proof. As an intermediate step, let $a=x^2,b=-y^2$. The first system then becomes:
$$\left\{\begin{array}{c}
u=a+b \\
-v^2=ab
\end{array}\right.$$
The equation $z^2-uz-v^2=0$ has solutions $a,b$, but the solutions can also be expressed as $\frac{u+\sqrt{u^2+4v^2}}{2}$ and $\frac{u-\sqrt{u^2+4v^2}}{2}$, by the solution formula for second-degree equations. It would seem we have an ambiguity, but in fact we don't. Because remember $a,b$ are expressed in terms of $x^2,y^2$, so we know their signs. In particular, $a=x^2>0$ and $b=-y^2<0$, so $a$ must be the one with the plus, and $b$ the one with the minus, since the one with the minus has $u-\sqrt{u^2+4v^2}<0$ as a numerator. This inequality is proven by placing the root on the right side, noting that if $u<0$ the inequality is obvious, and squaring to produce the obvious inequality $4v^2\geq0$ in the $u>0$ case. Thus the above system has yielded:
$$\left\{\begin{array}{c}
a=\frac{u+\sqrt{u^2+4v^2}}{2} \\
b=\frac{u-\sqrt{u^2+4v^2}}{2}
\end{array}\right.,$$
which implies:
$$\left\{\begin{array}{c}
x^2=a=\frac{u+\sqrt{u^2+4v^2}}{2} \\
y^2=-b=\frac{\sqrt{u^2+4v^2}-u}{2}
\end{array}\right.,$$
which is exactly the second system on top, provided that $x,y$ are positive as in the hypothesis of the question. $\square$
Alternate proof. Compute $u(x(u,v),y(u,v))$ and $v(x(u,v),y(u,v))$, i.e. shove those expressions of $x,y$ into $u,v$, and reach the identity. This proves $\Leftarrow$. Shoving $u,v$ into $x,y$ gives $x=|x|,y=|y|$, thus proving $\Rightarrow$ when $x,y>0$. $\square$
OP-suggested strategy
As suggested by the OP in a comment to this answer, one could try assuming $u_1=u_2,v_1=v_2$ and seeing what this implies. Let us write the two equations in terms of $x_1,x_2,y_1,y_2$:
$$\left\{\begin{array}{c}
x_1^2-y_1^2=x_2^2-y_2^2 \\
x_1y_1=x_2y_2
\end{array}\right.$$
Assuming $y_1\neq0$, we can divide the second equation, getting to:
$$x_1=\frac{x_2y_2}{y_1}.$$
Substituting this into equation 1 yields:
$$\frac{x_2^2y_2^2}{y_1^2}-y_1^2=x_2^2-y_2^2\iff x_2^2y_2^2-y_1^2y_1^2=x_2^2y_1^2-y_2^2y_1^2\iff x_2^2(y_2^2-y_1^2)=-y_1^2(y_2^2-y_1^2),$$
which in real numbers necessarily implies $y_2^2-y_1^2=0$, i.e. $y_1=\pm y_2$. So let us rewrite the system to then substitute this into equation two:
$$\left\{\begin{array}{c}
y_1=\pm y_2 \\
x_1=\frac{x_2y_2}{y_1} \\
y_1\neq0
\end{array}\right.
\iff
\left\{\begin{array}{c}
y_1=\pm y_2 \\
x_1=\pm x_2 \\
y_1\neq0
\end{array}\right.$$
Adding the hypothesis that the $x$'s and $y$'s be positive (or nonnegative) rules out the $\pm$s, yielding $x_1=x_2,y_1=y_2$. Of course, we still have the $y_1=0$ case to analyze. Let us add this hypothesis to the starting system and rewrite it accordingly:
$$\left\{\begin{array}{c}
x_1^2=x_2^2-y_2^2 \\
0=x_2y_2
\end{array}\right.$$
Equation two yields $x_2=0$ or $y_2=0$. Substituting $x_2=0$ into equation 1 yields $x_1^2=-y_2^2$, which has no real solutions. Thus, the hypothesis $y_1=0$ in the real field implies $y_2=0$, so $y_1=y_2$. The first equation then reduces to $x_1^2=y_1^2\iff x_1=\pm x_2$, which again is turned to $x_1=x_2$ by the sign hypothesis, proving injectivity. Of course, finding the inverse can be handy, but it is not necessary to prove injectivity in this case.
