This homework exercise comes from M. Boas, Mathematical Methods in the Physical Sciences:
Typed out: If $xs^2+yt^2=1$ and $x^s+y^2t=xy-4$, find $\partial x/\partial s$, $\partial x/\partial t$, $\partial y/\partial s$, $\partial y/\partial t$, at $(x,y,s,t)=(1,-3,2,-1)$. Hint: To simplify the work, substitute the numerical values just after you have taken differentials.
So, if I understand the question right there are two shapes, given by the two equations, and they intersect somewhere in 4-D space. We want to find just how much e.g. d$x$ changes when we change d$s$ while keeping $y$ and $t$ fixed.
The differentials I find to be: \begin{align} s^2 (dx)+2xs(ds) + 2yt(dt) + t^2(dy) & = 0,\\ 2sx(dx) + x^2 (ds) + y^2(dt)+2ty(dy) & = x(dy) + y(dx). \end{align}
I substitute the given values and simplify, \begin{align} 4(dx) + 4(ds) + 6(dt) + (dy) & = 0,\\ 4(dx) + (ds) + 9(dt) + 5(dy) & = 0. \end{align}
Initially I assumed that, to find $\partial x/\partial s$, I could just set $(dt)=(dy)=0$, but that system doesn't have a solution. So it appears as if I need to solve the system fully... Using Cramer's rule I obtain
\begin{align}
(dx) = \frac{\left\lvert\begin{array}{cc} -6(dt)-(dy) & 4 \\ -9(dt)-5(dy)& 1 \end{array} \right\rvert}{
\left\lvert \begin{array}{cc} 4 & 4 \\ 7 & 1 \end{array} \right\rvert } = \frac{30(dt)+19(dy)}{-24}.
\end{align}
Similarly for $(ds)$ I obtain
\begin{align}
(ds) = \frac{\left\lvert\begin{array}{cc} 4 & -6(dt)-(dy) \\ 7 & -9(dt)-5(dy) \end{array} \right\rvert}{
\left\lvert \begin{array}{cc} 4 & 4 \\ 7 & 1 \end{array} \right\rvert } = \frac{6(dt)-13(dy)}{-24}.
\end{align}
Taking the ratio of the two gives
$$
\frac{dx}{ds}=\frac{30(dt)+19(dy)}{6(dt)-13(dy)}
$$
The correct answer, according to the answer sheet, is $\partial x/\partial s=-19/13$.
I repeated the procedure for all the four desired partial derivatives, to find: \begin{align} \frac{dx}{ds}&=\frac{30(dt)+19(dy)}{6(dt)-13(dy)},\\ \frac{dx}{dt}&=\frac{-30(ds)+21(dy)}{24(ds)-13(dy)},\\ \frac{dy}{ds}&=\frac{-24(dx)+2(dt)}{-13(dx)-21(dt)},\\ \frac{dy}{dt}&=\frac{-6(dx)+30(ds)}{-13(dx)-19(ds)}. \end{align} Comparing to the answers, I would obtain the right solution if I would set $(ds)$ or $(dt)$ to zero on the right-hand side each time. But I don't see anything in the question that motivatives why such an assumption is appropriate. Was there anything in the question/equations that suggests that I could set $(dt)$ or $(ds)$ to zero? Did I take the wrong route to obtain my answers?
Either bad notations, or the books explained the notations elsewhere. This is why partial derivative is confusing, especially when used in physics without clear explanation. The simple reason is that $x,y$ are being treated as a functions in term of $s,t$ only. So the partial derivative? They are keeping only one other variable constant. For example $\frac{\partial x}{\partial s}$ keep $t$ constant, so you can only set $dt=0$, $dy$ is not affected.
Which makes sense. $2$ equations in $4$-dimensional space should cut out a surface, which is $2$-dimensional, so you can only use $2$ variables as coordinate.