Suppose a non-identically zero function $f$ satisfies $f\left(x\right)f\left(y\right)=f\left(\sqrt{x^2+y^2}\right)$ for all $x$ and $y$. Compute $$f\left(1\right)-f\left(0\right)-f\left(-1\right).$$
1. Given the functional equation \( f(x)f(y) = f(\sqrt{x^2 + y^2}) \) for all \( x \) and \( y \), we need to find \( f(1) - f(0) - f(-1) \).

2. First, let's consider the case when \( x = 0 \):
   \[
   f(0)f(y) = f(\sqrt{0^2 + y^2}) = f(|y|)
   \]
   This implies:
   \[
   f(0)f(y) = f(y) \quad \text{for all } y
   \]
   Since \( f \) is not identically zero, there exists some \( y \) such that \( f(y) \neq 0 \). For such \( y \), we can divide both sides by \( f(y) \):
   \[
   f(0) = 1
   \]

3. Next, consider the case when \( y = 0 \):
   \[
   f(x)f(0) = f(\sqrt{x^2 + 0^2}) = f(|x|)
   \]
   Since \( f(0) = 1 \), we have:
   \[
   f(x) = f(|x|)
   \]
   This implies that \( f \) is an even function, i.e., \( f(x) = f(-x) \).

4. Now, consider the case when \( x = y = 1 \):
   \[
   f(1)f(1) = f(\sqrt{1^2 + 1^2}) = f(\sqrt{2})
   \]
   This simplifies to:
   \[
   f(1)^2 = f(\sqrt{2})
   \]

5. Consider the case when \( x = 1 \) and \( y = -1 \):
   \[
   f(1)f(-1) = f(\sqrt{1^2 + (-1)^2}) = f(\sqrt{2})
   \]
   Since \( f \) is even, \( f(-1) = f(1) \), so:
   \[
   f(1)f(1) = f(\sqrt{2})
   \]
   This simplifies to:
   \[
   f(1)^2 = f(\sqrt{2})
   \]
   This is consistent with the previous result.

6. From the above, we have \( f(1)^2 = f(\sqrt{2}) \). Since \( f \) is not identically zero, \( f(1) \neq 0 \). Let \( f(1) = c \). Then:
   \[
   c^2 = f(\sqrt{2})
   \]

7. Finally, we need to compute \( f(1) - f(0) - f(-1) \):
   \[
   f(1) - f(0) - f(-1) = c - 1 - c = -1
   \]

The final answer is \(\boxed{-1}\).