How to find $Z_y(x_0,y_0)$ if $Z(x, y) = Z(x(u,v), y(u,v)) =T(u,v)$ and $u=xy$, $v={y\over x}$?

Question

Calculate $Z_y(2,1.5)$ if $Z(x, y) = Z(x(u,v), y(u,v)) =T(u,v)$ and $u=xy$, $v={y\over x}$.

I think that according to chain rule: $$ T_u=Z_xx_u+Z_yy_u\\ T_v=Z_xx_v+Z_yy_v $$ Using implicit differentiation and product rule: $$ 1=x_uy+xy_u\\ 1=\frac{x_vy-xy_v}{y^2}\\ 0=x_vy+xy_v\\ 0=\frac{x_uy-xy_u}{y^2} $$ which yields: $$ x_v=y\\ y_v=-\frac{y}{2x} $$ So at point $(2,1.5)$ we have $T_v=1.5Z_x+0.375Z_y$.

But I can't progress further because when trying to calculate $x_u$: $$ x_u=\frac{1+xy_u}{y} \quad\text{from 1st equation}\\ \text{plugging this into the last equation:}\quad 0=\frac{1}{y}\implies y=0 $$

Answer 1

$\textbf{Hint}$: Use the chain rule.

\begin{align*} T(u,v)=T(u(x,y),v(x,y)) \Rightarrow Z_y = \frac{\partial T}{\partial y} = \begin{pmatrix} T_u & T_v \end{pmatrix} \cdot \begin{pmatrix} u_y \\ v_y\end{pmatrix} \end{align*}

Answer 2

Using differentials: $$dZ=dT=T_u du+T_v dv, $$ and also $$du=u_x d x+u_y dy, \\ dv=v_x dx+v_ydy .$$ Thus $$dZ=(T_u u_x+T_v v_x) dx+(T_uu_y+T_v v_y)dy,$$ which gives $$Z_y(x_0,y_0)=x_0T_u+\frac{T_v}{x_0}. $$ The partial derivatives $T_u,T_v$ are calculated at the point $(u_0,v_0)$ which corresponds to $(x_0,y_0)$.