Answer
See below
Work Step by Step
The transpose of the given matrix is:
$A^T=\frac{1}{1+x^2}\begin{bmatrix}
1& 2x &2x^2\\-2x &1-2x^2&2x\\2x^2 & -2x &1
\end{bmatrix}$
Find the product of matrix $A$ and its transpose:
$A.A^T=\frac{1}{1+x^2}\begin{bmatrix}
1& -2x &2x^2\\2x &1-2x^2&-2x\\2x^2 & 2x &1
\end{bmatrix}\frac{1}{1+x^2}\begin{bmatrix}
1& 2x &2x^2\\-2x &1-2x^2&2x\\2x^2 & -2x &1
\end{bmatrix}\\
=\frac{1}{1+x^2}\frac{1}{1+x^2}\begin{bmatrix}
1& -2x &2x^2\\-2x &1-2x^2&-2x\\2x^2 & 2x &1
\end{bmatrix}\begin{bmatrix}
1& 2x &2x^2\\-2x &1-2x^2&2x\\2x^2 & -2x &1
\end{bmatrix}\\
=\frac{1}{1+x^2}\begin{bmatrix}
1+4x +4x^2 & 0 & 0\\0 &1+4x +4x^2&0\\0 & 0 &1+4X+4x^2
\end{bmatrix}\\
=\begin{bmatrix}
1 & 0 & 0\\0 &1&0\\0 & 0 &12
\end{bmatrix}$
We can see that $A^T$ is inverse of the matrix $A$. Hence, matrix $A$ is orthogonal.
