Explanation:

To determine the stability of a function, we need to analyze its behavior over time. In this case, the given function is a difference equation of the form y[n] = y[n-1] * x[n].

Definition of Stability:
A system is said to be stable if its output remains bounded for any bounded input. In other words, a stable system does not exhibit unbounded growth or oscillations in its output.

Analyzing the Given Function:
To analyze the stability of the given function, let's consider a bounded input signal x[n] = A, where A is a constant.

Case 1: A > 0
If A is positive, the function becomes y[n] = y[n-1] * A.

For the initial condition y[-1] = y[n0], where n0 is a particular time point,
y[0] = y[-1] * A = y[n0] * A
y[1] = y[0] * A = y[n0] * A^2
y[2] = y[1] * A = y[n0] * A^3
...
y[n] = y[n-1] * A = y[n0] * A^(n+1)

As n increases, the value of y[n] grows exponentially with A^(n+1). Hence, the output is unbounded and the system is unstable.

Case 2: A = 0
If A is zero, the output y[n] will be zero for all values of n. Hence, the system is stable as the output remains bounded.

Case 3: A < />
If A is negative, the function becomes y[n] = -y[n-1] * |A|.

For the initial condition y[-1] = y[n0], where n0 is a particular time point,
y[0] = -y[-1] * |A| = -y[n0] * |A|
y[1] = -y[0] * |A| = y[n0] * |A|^2
y[2] = -y[1] * |A| = -y[n0] * |A|^3
...
y[n] = -y[n-1] * |A| = -y[n0] * |A|^(n+1)

Similar to Case 1, as n increases, the value of y[n] grows exponentially with |A|^(n+1). Hence, the output is unbounded and the system is unstable.

Conclusion:
From the analysis of the different cases, we can see that the given function y[n] = y[n-1] * x[n] is stable only when the input signal x[n] is zero (i.e., A = 0). In all other cases, the output grows exponentially and the system is unstable. Therefore, the correct answer is option 'A' - It is stable.