Showing the existance of the functions $u,v,s,t$ by Implicit function theorem. I am studying from William Wade's introduction to analysis book the question 11.6.5 at page 434 
Question:
The given nonzero numbers $x_0, y_0, u_0, v_0, s_0, t_0$ which is simultanuously equations
(*)$$u^2+sx+ty=0$$
$$v^2+tx+sy=0$$
$$2s^2x+2t^2y-1=0$$
$$s^2x-t^2y=0$$
Prove that there exist the functions $u(x,y), v(x,y), s(x,y), t(x,y)$ and an open ball $B$ containing $(x_0,y_0)$ such that $u,V,s,t$ are continuously differentiable and satisfies (*) on $B$ and such that $u(x_0,y_0)=u_0, v(x_0,y0_)=v_0, s(x_0,y_0)=s_0, t(x_0,y_0)=t_0$

My solution trial:
Let's define a function 
$$F(x,y,u,v,s,t)=(u^2+sx+ty,v^2+tx+sy, 2s^2x+2t^2y-1,s^2x-t^2y)$$
Let's find the determinant of $F$
$$DF(x,y,u,v,s,t)= \begin{pmatrix} s & t & 2s^2 & s^2 \\ t & s & 2t^2 & -t^2 \\ 2u & 0& 0&0 \\ 0 & 2v & 0&0 \\ x & y & 4sx & 2sx \\ y & x& 4ty & -2ty\end{pmatrix}$$
Since $x,y,s,t,u,v$ are nonzero, $DF\not=0$
Since $Daf$ is nonzero, I can apply to Implicit function theorem so as to show that functions $u,V,s,t$ exists.
But how apply this theorem? Pleae can someone explain This last sentence more? I thought  there but I cannot apply the theorem. Thank you for helping:) 
 A: Write your $F$ function as
$$
F(x,y,u,v,s,t)=(F_1(x,y,u,v,s,t), F_2(x,y,u,v,s,t), F_3(x,y,u,v,s,t),F_4(x,y,u,v,s,t))
$$
You must show that the matrix
$$
\left(\begin{array}{llll}
\dfrac{\partial F_1}{\partial u}&\dfrac{\partial F_2}{\partial u}\dfrac{\partial F_3}{\partial s}&\dfrac{\partial F_4}{\partial u}\\
\dfrac{\partial F_1}{\partial v}&\dfrac{\partial F_2}{\partial v}\dfrac{\partial F_3}{\partial s}&\dfrac{\partial F_4}{\partial v}\\
\dfrac{\partial F_1}{\partial s}&\dfrac{\partial F_2}{\partial s}\dfrac{\partial F_3}{\partial s}&\dfrac{\partial F_4}{\partial s}\\
\dfrac{\partial F_1}{\partial t}&\dfrac{\partial F_2}{\partial t}\dfrac{\partial F_3}{\partial s}&\dfrac{\partial F_4}{\partial t}\\
\end{array}\right)
$$
has non-zero determinant, that is. This is the necessary condition for the Implicit Function Theorem, and guarantees that there is a neigborhood U of $(x_0,y_0)$ where you can write
$$
u=u(x,y),v=v(x,y),s=s(x,y),t=t(x,y)
$$
such that
$$
F(x,y,u(x,y),v(x,y),s(x,y),t(x,y))=0
$$

Notice that this is a submatrix of the Jacobian matrix you wrote in your question. The trick of the Implicit Function Theorem is, if your function is $F:\mathbb{R}^{m+n}\rightarrow\mathbb{R}^m$, finding the $m\times m$ submatrix of the Jacobian matrix of $F$ (which is $m\times (m+n)$) that corresponds the derivations of the $m$ $F_i$ functions with respect to the variables that you want to become functions (in your case, $(u,v,s,t)$), and looking if it has non-zero determinant.
Phew!
