Multivariable calculus - partial derivative I got this question:

Let $z=\frac{2x}{y}+\ln{(xy)}$ where $x$ and $y$ are functions of $s$ and $t$ defined as $s=x-y^2$ and $t=x^2+y$.
Find $\frac{\partial z}{\partial s}$ when $x=1$ and $y=1$.

Does this question make sense? To me it seems quite ill-posed, I can't figure out how to apply the usual chain rule (is it even possible to do that?)
 A: I'm going to take this opportunity to introduce a non$-$conventional notation for partial derivatives I learned while tutoring a student attending MIT. This notation may clear up some of the confusion you're having when the variables are all jumbled up like they are here.
You're looking to compute $\Big(\frac{\partial z}{\partial s}\Big)_t$ which in words represents "the partial derivative of $z$ with respect to $s$ while keeping $t$ fixed." From chain rule we have $$\Big(\frac{\partial z}{\partial s}\Big)_t=z_x\Big(\frac{\partial x}{\partial s}\Big)_t+z_y\Big(\frac{\partial y}{\partial s}\Big)_t$$ $$\Big(\frac{\partial s}{\partial s}\Big)_t=\Big(\frac{\partial x}{\partial s}\Big)_t-2y\Big(\frac{\partial y}{\partial s}\Big)_t$$ $$\Big(\frac{\partial t}{\partial s}\Big)_t=2x\Big(\frac{\partial x}{\partial s}\Big)_t+\Big(\frac{\partial y}{\partial s}\Big)_t$$ Since $\Big(\frac{\partial s}{\partial s}\Big)_t=1$ and $\Big(\frac{\partial t}{\partial s}\Big)_t=0$ we have that $$\Big(\frac{\partial z}{\partial s}\Big)_t=z_x\Big(\frac{\partial x}{\partial s}\Big)_t+z_y\Big(\frac{\partial y}{\partial s}\Big)_t$$ $$1=\Big(\frac{\partial x}{\partial s}\Big)_t-2y\Big(\frac{\partial y}{\partial s}\Big)_t$$ $$0=2x\Big(\frac{\partial x}{\partial s}\Big)_t+\Big(\frac{\partial y}{\partial s}\Big)_t$$ As I said in my comments, the last two equations provide us with a system which we can solve for $\Big(\frac{\partial x}{\partial s}\Big)_t$ and $\Big(\frac{\partial y}{\partial s}\Big)_t$ explicitly. Doing so gives us an expression for $\Big(\frac{\partial z}{\partial s}\Big)_t$: $$\Big(\frac{\partial z}{\partial s}\Big)_t=\frac{z_x-2xz_y}{4xy+1}$$ You might ask: How would we compute something like $\Big(\frac{\partial z}{\partial s}\Big)_x$? In words this represents "the partial derivative of $z$ with respect to $s$ while keeping $x$ fixed." Chain rule gives us a similar set of equations to work with: $$\Big(\frac{\partial z}{\partial s}\Big)_x=z_x\Big(\frac{\partial x}{\partial s}\Big)_x+z_y\Big(\frac{\partial y}{\partial s}\Big)_x$$ $$\Big(\frac{\partial s}{\partial s}\Big)_x=\Big(\frac{\partial x}{\partial s}\Big)_x-2y\Big(\frac{\partial y}{\partial s}\Big)_x$$ $$\Big(\frac{\partial t}{\partial s}\Big)_x=2x\Big(\frac{\partial x}{\partial s}\Big)_x+\Big(\frac{\partial y}{\partial s}\Big)_x$$ Because $\Big(\frac{\partial s}{\partial s}\Big)_x=1$ and $\Big(\frac{\partial x}{\partial s}\Big)_x=0$ the three equations above boil down to $$\Big(\frac{\partial z}{\partial s}\Big)_x=z_y\Big(\frac{\partial y}{\partial s}\Big)_x$$ $$1=-2y\Big(\frac{\partial y}{\partial s}\Big)_x$$ $$\Big(\frac{\partial t}{\partial s}\Big)_x=\Big(\frac{\partial y}{\partial s}\Big)_x$$ Combining the first two give $\Big(\frac{\partial z}{\partial s}\Big)_x=-\frac{z_y}{2y}$. For practice, see if you can compute $\Big(\frac{\partial z}{\partial y}\Big)_t$ and $\Big(\frac{\partial z}{\partial t}\Big)_x$. This notation does take some getting used to, but if you can get past its ugliness, I guarantee you will find it helpful.
