Assume that \( F(x, y, z(x, y))=0 \) implicitly defines \( z \) as a differentiable function of \( x \) and \( y \). The partial derivatives of \( z \) are \( \frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}} \) and \( \frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}} \). Evaluate \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) for \( x y z+2 x+3 y-5 z=0 \). \( \frac{\partial z}{\partial x}=\square \)

To find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for the function defined by \(F(x, y, z) = xy z + 2x + 3y - 5z = 0\), we first need to compute the necessary partial derivatives of \(F\). 1. **Calculate the required partial derivatives:** - \( F_x = yz + 2 \) - \( F_y = xz + 3 \) - \( F_z = xy - 5 \) 2. **Evaluate \(\frac{\partial z}{\partial x}\):** Using the formula \(\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}\), \[ \frac{\partial z}{\partial x} = -\frac{yz + 2}{xy - 5}. \] Thus, \[ \frac{\partial z}{\partial x} = -\frac{yz + 2}{xy - 5} = \square. \] Now let’s find \(\frac{\partial z}{\partial y}\). 3. **Evaluate \(\frac{\partial z}{\partial y}\):** Using the formula \(\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}\), \[ \frac{\partial z}{\partial y} = -\frac{xz + 3}{xy - 5}. \] So, \[ \frac{\partial z}{\partial y} = -\frac{xz + 3}{xy - 5}. \]