Subtract $ \dfrac{k}{n-k} $ from $ \dfrac{k}{k-n} $ to get $ \dfrac{ \color{purple}{ -2k^2+2kn } }{ -k^2+2kn-n^2 }$.
To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ -k+n }$ and the second by $\color{blue}{ k-n }$.
$$ \begin{aligned} \frac{k}{k-n} - \frac{k}{n-k} & = \frac{ k \cdot \color{blue}{ \left( -k+n \right) }}{ \left( k-n \right) \cdot \color{blue}{ \left( -k+n \right) }} -
\frac{ k \cdot \color{blue}{ \left( k-n \right) }}{ \left( n-k \right) \cdot \color{blue}{ \left( k-n \right) }} = \\[1ex] &=\frac{ \color{purple}{ -k^2+kn } }{ -k^2+kn+kn-n^2 } - \frac{ \color{purple}{ k^2-kn } }{ -k^2+kn+kn-n^2 } = \\[1ex] &=\frac{ \color{purple}{ -k^2+kn - \left( k^2-kn \right) } }{ -k^2+2kn-n^2 }=\frac{ \color{purple}{ -2k^2+2kn } }{ -k^2+2kn-n^2 } \end{aligned} $$