Find the following derivatives. Express your answer in terms of the independent variables. \[ w_{s} \text { and } w_{t} \text {, where } w=\frac{x-z}{3 y+z}, x=3 s+4 t, y=s t \text {, and } z=3 s-4 t \] \( \frac{\partial w}{\partial x}=\frac{1}{3 y+z} \) (Type an expression using \( x, y \), and \( z \) as the variables.) \( \frac{\partial w}{\partial y}=\square \) (Type an expression using \( x, y \), and \( z \) as the variables.)

To find the derivatives \( w_s \) and \( w_t \), we will first express \( w = \frac{x-z}{3y+z} \) in the context of the independent variables. The first derivative is given: \[ \frac{\partial w}{\partial x} = \frac{1}{3y + z} \] Now we can calculate \( \frac{\partial w}{\partial y} \): We want to apply the quotient rule: \[ \frac{\partial w}{\partial y} = \frac{(3y + z) \frac{\partial (x-z)}{\partial y} - (x-z) \frac{\partial (3y + z)}{\partial y}}{(3y + z)^2} \] Calculating the necessary partial derivatives: - \( \frac{\partial (x - z)}{\partial y} = 0 \) (since \( x \) and \( z \) do not depend on \( y \)), - \( \frac{\partial (3y + z)}{\partial y} = 3 \). So, substituting these into the quotient rule gives: \[ \frac{\partial w}{\partial y} = \frac{(3y + z) \cdot 0 - (x - z) \cdot 3}{(3y + z)^2} = \frac{-3(x - z)}{(3y + z)^2} \] Thus, we have: \[ \frac{\partial w}{\partial y} = \frac{-3(x - z)}{(3y + z)^2} \] Finally, here are the answers: \[ \frac{\partial w}{\partial y} = \frac{-3(x - z)}{(3y + z)^2} \]