This is Example C in section 1.1 from Advanced Calculus (2nd ed) by David Widder.
$$\begin{cases} v + \log u = xy \\ u + \log v = x - y \end{cases}$$
$$\begin{cases} \frac{1}{u} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial x} = y \\ \frac{1}{v} \frac{\partial v}{\partial x} + \frac{\partial u}{\partial x} = 1 \end{cases}$$
$$\frac{\partial u}{\partial v} = \frac{\begin{vmatrix} yu & u \\ v & 1 \end{vmatrix}}{\begin{vmatrix} 1 & u \\ v & 1 \end{vmatrix}} = \frac{u(y-v)}{1-uv}$$
How does the author come up with the ratio of determinants in the third equation?
Use implicit differentiation of $v + \ln u = x y$ and $u + \ln v = x - y$ to get expressions for $\dfrac {\partial u} {\partial x}$ and $\dfrac {\partial v} {\partial x}$.
You should get: $$\dfrac {\partial u} {\partial x} + u \dfrac {\partial v} {\partial x} = y u$$
and: $$v \dfrac {\partial u} {\partial x} + \dfrac {\partial v} {\partial x} = v$$
Then you can combine the above into simultaneous linear equations expressed in matrix form as:
$$\begin {pmatrix} 1 & u \\ v & 1 \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial x} \\ \dfrac {\partial v} {\partial x} \end {pmatrix} = \begin {pmatrix} y u \\ v \end {pmatrix}$$
and then it's a matter of using Cramer's Rule to evaluate:
$$\dfrac {\partial u} {\partial x} = \dfrac {\begin {vmatrix} y u & u \\ v & 1 \end {vmatrix} } {\begin {vmatrix} 1 & u \\ v & 1 \end {vmatrix} } = \dfrac {u (y - v) } {1 - u v}$$
and:
$$\dfrac {\partial v} {\partial x} = \dfrac {\begin {vmatrix} 1 & y u \\ v & v \end {vmatrix} } {\begin {vmatrix} 1 & u \\ v & 1 \end {vmatrix} } = \dfrac {v (1 - y u) } {1 - u v}$$
exactly as though this were a pair of simultaneous equations.
This of course is done (either explicitly or implicitly) by taking the inverse of the matrices $\begin {pmatrix} 1 & u \\ v & 1 \end {pmatrix}$ and $\begin {pmatrix} y u & u \\ v & 1 \end {pmatrix}$ and $\begin {pmatrix} 1 & y u \\ v & v \end {pmatrix}$.
This involves calculating their determinants.