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 $$ x y z+2 x+3 y-5 z=0 $$. $$ \frac{\partial z}{\partial x}=-\frac{y z+2}{x y-5} $$ $$ \frac{\partial z}{\partial y}= $$

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To evaluate the partial derivatives $$ \frac{\partial z}{\partial x} $$ and $$ \frac{\partial z}{\partial y} $$ for the equation $$ F(x, y, z) = xy z + 2x + 3y - 5z = 0 $$, we first need to compute the necessary partial derivatives of $$ F $$.

1. **Calculate the partial derivatives of $$ F $$**:
   - $$ F_x = \frac{\partial F}{\partial x} $$
   - $$ F_y = \frac{\partial F}{\partial y} $$
   - $$ F_z = \frac{\partial F}{\partial z} $$

Let's compute these derivatives step by step.

### Step 1: Compute $$ F_x $$, $$ F_y $$, and $$ F_z $$

- $$ F(x, y, z) = xy z + 2x + 3y - 5z $$

**Partial derivative with respect to $$ x $$**:
$$
F_x = y z + 2
$$

**Partial derivative with respect to $$ y $$**:
$$
F_y = x z + 3
$$

**Partial derivative with respect to $$ z $$**:
$$
F_z = xy - 5
$$

### Step 2: Substitute into the formulas for $$ \frac{\partial z}{\partial x} $$ and $$ \frac{\partial z}{\partial y} $$

Now we can use the formulas for the partial derivatives of $$ z $$:
$$
\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}
$$
$$
\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}
$$

### Step 3: Calculate $$ \frac{\partial z}{\partial x} $$

Substituting $$ F_x $$ and $$ F_z $$:
$$
\frac{\partial z}{\partial x} = -\frac{y z + 2}{xy - 5}
$$

This matches the given expression for $$ \frac{\partial z}{\partial x} $$.

### Step 4: Calculate $$ \frac{\partial z}{\partial y} $$

Now substituting $$ F_y $$ and $$ F_z $$:
$$
\frac{\partial z}{\partial y} = -\frac{x z + 3}{xy - 5}
$$

### Final Results

Thus, the evaluated partial derivatives are:
$$
\frac{\partial z}{\partial x} = -\frac{y z + 2}{xy - 5}
$$
$$
\frac{\partial z}{\partial y} = -\frac{x z + 3}{xy - 5}
$$
$$ \frac{\partial z}{\partial x} = -\frac{y z + 2}{xy - 5} $$ $$ \frac{\partial z}{\partial y} = -\frac{x z + 3}{xy - 5} $$

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