The full question is Q6.86 in the picture.
How do you prove, $f(x,y,r,s)=0,g(x,y,r,s)=0, then \frac{\partial y}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial y}{\partial s}\frac{\partial s}{\partial x}=0?$.
I got, $$df=f_xdx=f_ydy+f_rdr+f_sds=0$$ $$dg=g_xdx=g_ydy+g_rdr+g_sds=0$$
Letting x and y be a function of r and s,
$$dx=x_rdr+x_Sds, dy=y_rdr+y_Sds$$
After subbing in dx and dy into the above equations, I do not have any idea how to continue.
Thanks in advance.
After some thinking I finally came up with a solution to the question. I've decided to post it here in case anyone else wants the answer.
$$F(x,y,r,s)=0, G(x,y,r,s)=0$$ $$dF=F_xdx+F_ydy+F_rdr+F_Sds=0$$ $$dG=G_xdx+G_ydy+G_rdr+G_Sds=0$$
Working out dy/dr, x and y are functions of r and s $$F_x\frac{dx}{dr}+F_y\frac{dy}{dr}+F_r\frac{dr}{dr}+F_S\frac{ds}{dr}=0$$ $$G_x\frac{dx}{dr}+G_y\frac{dy}{dr}+G_r\frac{dr}{dr}+G_S\frac{ds}{dr}=0$$
As s and r are independent and dr/dr=1,
$$F_x\frac{dx}{dr}+F_y\frac{dy}{dr}+F_r=0$$ $$G_x\frac{dx}{dr}+G_y\frac{dy}{dr}+G_r=0$$ $$F_x\frac{dx}{dr}+F_y\frac{dy}{dr}=-F_r$$ $$G_x\frac{dx}{dr}+G_y\frac{dy}{dr}=-G_r$$
$$ \frac{dy}{dr}=\frac{ \begin{vmatrix} F_x& -F_r\\ G_x & -G_r \\ \end{vmatrix}}{\begin{vmatrix} F_x& F_y\\ G_x & G_y \\ \end{vmatrix}} $$
For $\frac{dy}{ds}$, $$F_x\frac{dx}{ds}+F_y\frac{dy}{ds}=-F_s$$ $$G_x\frac{dx}{ds}+G_y\frac{dy}{ds}=-G_s$$
The arguments to obtain the above 2 equations are simply the same as when we obtained $\frac{dy}{dr}$.
Solving,
$$ \frac{dy}{ds}=\frac{ \begin{vmatrix} F_x& -F_s\\ G_x & -G_s \\ \end{vmatrix}}{\begin{vmatrix} F_x& F_y\\ G_x & G_y \\ \end{vmatrix}} $$
Now, by the Implicit Function Theorem, we know that we can express r and s as a function of x and y. $r=r'(x,y),s=s'(x,y)$ so as long as the determinant does not vanish. So we shall keep that as a condition without loss of generality.
Continuing, we implicitly differentiate with respect to $dx$,
$$F_r\frac{dr}{dx}+F_s\frac{ds}{dx}=-F_x$$ $$G_r\frac{dr}{dx}+G_s\frac{ds}{dx}=-G_x$$
Solving, we already have $\frac{dr}{dx}, \frac{ds}{dx}$
$$ \frac{dr}{dx}=\frac{ \begin{vmatrix} -F_x& F_s\\ -G_x & G_s \\ \end{vmatrix}}{\begin{vmatrix} F_r& F_s\\ G_r & G_s \\ \end{vmatrix}} $$
$$ \frac{ds}{dx}=\frac{ \begin{vmatrix} F_r& -F_x\\ G_r & -G_x \\ \end{vmatrix}}{\begin{vmatrix} F_s& F_s\\ G_r & G_s \\ \end{vmatrix}} $$
Hence, we have $\frac{dy}{dr}\frac{dr}{dx}$ is equal to,
$$ \frac{ \begin{vmatrix} F_x& -F_r\\ G_x & -G_r \\ \end{vmatrix}}{\begin{vmatrix} F_x& F_y\\ G_x & G_y \\ \end{vmatrix}}.\frac{ \begin{vmatrix} -F_x& F_s\\ -G_x & G_s \\ \end{vmatrix}}{\begin{vmatrix} F_r& F_s\\ G_r & G_s \\ \end{vmatrix}}=\frac{**F_x(-G_r)+G_xF_r**}{F_xG_y-G_xF_y}.\frac{-F_xG_s+G_xF_s}{F_rG_s-G_rF_s} $$
Likewise, for $\frac{dy}{dr}\frac{dr}{dx}$ $$ \frac{ \begin{vmatrix} F_x& -F_s\\ G_x & -G_s \\ \end{vmatrix}}{\begin{vmatrix} F_x& F_y\\ G_x & G_y \\ \end{vmatrix}}.\frac{ \begin{vmatrix} F_r& -F_x\\ G_r & -G_x \\ \end{vmatrix}}{\begin{vmatrix} F_r& F_s\\ G_r & G_s \\ \end{vmatrix}}=\frac{F_x(-G_s)+G_xF_s}{F_xG_y-G_xF_y}.\frac{**-G_xF_r+G_rF_x**}{F_rG_s-G_rF_s} $$
Bolded terms are negative of each other, and from there it's trivally obvious that the theorem is true.