Answer: \(A=16\). The stationary points at \(x=-1\) and \(x=1\) are minimum points.

Let

\(u=\frac{x^2-1}{x^2+1}.\)

Then \(y=u^4\), so

\(\frac{\mathrm{d}y}{\mathrm{d}x}=4u^3\frac{\mathrm{d}u}{\mathrm{d}x}.\)

Using the quotient rule,

\(\frac{\mathrm{d}u}{\mathrm{d}x} =\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}.\)

The numerator simplifies to

\(2x\{(x^2+1)-(x^2-1)\}=4x.\)

Hence

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{4x}{(x^2+1)^2}.\)

Therefore

\(\frac{\mathrm{d}y}{\mathrm{d}x} =4\left(\frac{x^2-1}{x^2+1}\right)^3\frac{4x}{(x^2+1)^2} =\frac{16x(x^2-1)^3}{(x^2+1)^5}.\)

So \(A=16\).

Stationary points occur when

\(16x(x^2-1)^3=0.\)

Thus

\(x=0,\quad x^2-1=0,\)

so

\(x=-1,\quad x=0,\quad x=1.\)

The denominator is always positive, so the sign of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) depends on \(x(x^2-1)^3\).

For \(x\lt -1\), the derivative is negative. For \(-1\lt x\lt 0\), it is positive. Therefore \(x=-1\) is a minimum point.

For \(0\lt x\lt 1\), the derivative is negative, and for \(x\gt 1\), it is positive. Therefore \(x=1\) is also a minimum point.

So the two stationary points with the same nature are \(x=-1\) and \(x=1\), and both are minimum points.