Prove that there exist differential functions $x=x(s,t)$,$y=y(s,t)$ defined around the point $(2,-1)$ which satisfy $$xs^2+yt^2=1,x^2s+y^2t=xy-4$$ How many of these functions there are?
I've solved similar problems to this one where I had to prove in an equation of $x$ and $y$ I could write $y$ as a function of $x$. However I don't know how to solve this one.
You can treat $s,t$ as coefficients and $x,y$ as variables. You have the system of equations
\begin{cases} s^2 x + t^2 y = 1 \\ sx^2 + ty^2 = xy - 4 \end{cases}
Solving for $y$ in the first equation and plugging it into the second gives $$ sx^2 + t \left( \frac{1-s^2 x}{t^2} \right)^2 = x \left(\frac{1-s^2 x}{t^2}\right) - 4 $$
$$ st^3 x^2 + (1-s^2x)^2 = tx(1-s^2x) - 4t^3 $$ $$ (st^3 + s^4 + ts^2)x^2 - (2s^2+t)x + (1 + 4t^3) = 0 $$
This is a quadratic equation in $x$, whose solution exists if the determinant $$ D(s,t) = (2s^2+t)^2 - 4(st^3 + s^4 + ts^2)(1+4t^3) \ge 0 $$
This is true since $D(2,-1) > 0$, so we're done.
You can also conclude that there are two possible solutions for $x$, and likewise for $y$