To find the partial differential equation by eliminating the arbitrary functions $f$ and $g$ from the equation

$z = f(x^2 - y) + g(x^2 + y),$

we'll proceed with differentiation and elimination steps.

Step 1: Define the Variables for Partial Differentiation

Let $u = x^2 - y \quad \text{and} \quad v = x^2 + y.$

Then we can write $z = f(u) + g(v)$, where $f$ and $g$ are arbitrary functions of $u$ and $v$.

Step 2: Find Partial Derivatives with Respect to $x$ and $y$

1. Partial derivative with respect to $x$:

   Using the chain rule: $\frac{\partial z}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x} + g'(v) \cdot \frac{\partial v}{\partial x}.$

   Since $u = x^2 - y$ and $v = x^2 + y$, $\frac{\partial u}{\partial x} = 2x \quad \text{and} \quad \frac{\partial v}{\partial x} = 2x.$

   Therefore, $\frac{\partial z}{\partial x} = f'(u) \cdot 2x + g'(v) \cdot 2x = 2x \left( f'(u) + g'(v) \right).$

2. Partial derivative with respect to $y$:

   Similarly, $\frac{\partial z}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y} + g'(v) \cdot \frac{\partial v}{\partial y}.$

   Since $u = x^2 - y$ and $v = x^2 + y$, $\frac{\partial u}{\partial y} = -1 \quad \text{and} \quad \frac{\partial v}{\partial y} = 1.$

   Therefore, $\frac{\partial z}{\partial y} = f'(u) \cdot (-1) + g'(v) \cdot 1 = -f'(u) + g'(v).$

Step 3: Find the Second Order Mixed Partial Derivatives

1. Second partial derivative with respect to $x$:

   Differentiate $\frac{\partial z}{\partial x} = 2x (f'(u) + g'(v))$ with respect to $x$: $\frac{\partial^2 z}{\partial x^2} = 2(f'(u) + g'(v)) + 2x \left( f''(u) \cdot \frac{\partial u}{\partial x} + g''(v) \cdot \frac{\partial v}{\partial x} \right).$

   Substituting $\frac{\partial u}{\partial x} = 2x$ and $\frac{\partial v}{\partial x} = 2x$, $\frac{\partial^2 z}{\partial x^2} = 2(f'(u) + g'(v)) + 2x \left( f''(u) \cdot 2x + g''(v) \cdot 2x \right).$

   Simplifying further, $\frac{\partial^2 z}{\partial x^2} = 2(f'(u) + g'(v)) + 4x^2 (f''(u) + g''(v)).$

2. Second partial derivative with respect to $y$:

   Differentiate $\frac{\partial z}{\partial y} = -f'(u) + g'(v)$ with respect to $y$: $\frac{\partial^2 z}{\partial y^2} = -f''(u) \cdot \frac{\partial u}{\partial y} + g''(v) \cdot \frac{\partial v}{\partial y}.$

   Substituting $\frac{\partial u}{\partial y} = -1$ and $\frac{\partial v}{\partial y} = 1$, $\frac{\partial^2 z}{\partial y^2} = f''(u) + g''(v).$

Step 4: Derive the Partial Differential Equation

Now we can eliminate $f$ and $g$ by forming a relation among $\frac{\partial^2 z}{\partial x^2}$, $\frac{\partial^2 z}{\partial y^2}$, and $\frac{\partial z}{\partial x}$.

Notice that [ \frac{\partial^2 z}{\partial x^2} - 4x^2 \frac{\partial^2 z}{\partial y^2} = 2(f'(u) + g'(v)) - 4x^2 (f''(u) + g''(v)) +