We are tasked with proving the following equation:

$\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 = e^{-2s} \left[ \left( \frac{\partial f}{\partial s} \right)^2 + \left( \frac{\partial f}{\partial t} \right)^2 \right]$

where $z = f(x, y)$ and the variables $x$ and $y$ are expressed in terms of $s$ and $t$ as:

$x = e^s \cos t, \quad y = e^s \sin t$

Step-by-step Solution:

1. Chain rule for partial derivatives:

We begin by using the chain rule to express the derivatives of $f$ with respect to $x$ and $y$ in terms of the derivatives with respect to $s$ and $t$.

The chain rule for $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ gives:

$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial x}$ $\frac{\partial f}{\partial y} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial y} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial y}$

However, since $x = e^s \cos t$ and $y = e^s \sin t$, we directly differentiate these to simplify.

2. Find the partial derivatives of $x$ and $y$:

$\frac{\partial x}{\partial s} = e^s \cos t, \quad \frac{\partial x}{\partial t} = -e^s \sin t$ $\frac{\partial y}{\partial s} = e^s \sin t, \quad \frac{\partial y}{\partial t} = e^s \cos t$

3. Express derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$:

Using the chain rule, we can now compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ in terms of $s$ and $t$.

$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial s} \cdot \frac{\partial s}{\partial x} + \frac{\partial f}{\partial t} \cdot \frac{\partial t}{\partial x}$