Consider the matrix f(x) = [ cases x & - x & 0 x & x & 0 0 & 0 & 1...

Given: f\left(x\right) = \left[\begin{cases}\cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{cases}\right]

⇒ f\left(- x\right) = \left[\begin{cases}\cos x & \sin x & 0 \\ - \sin x & \cos x & 0 \\ 0 & 0 & 1\end{cases}\right]

⇒ f\left(x\right) \times f\left(- x\right) = \left[\begin{cases}\cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{cases}\right]\left[\begin{cases}\cos x & \sin x & 0 \\ - \sin x & \cos x & 0 \\ 0 & 0 & 1\end{cases}\right]

⇒ f\left(x\right) \times f\left(- x\right) = \left[\begin{cases}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}\right]

$⇒ f\left(x\right) \times f\left(- x\right) = I$

$⇒ f\left(- x\right) = {f}^{- 1}\left(x\right)$

Hence statement-I is correct

Now, checking statement II

⇒ f\left(y\right) = \left[\begin{cases}\cos y & - \sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{cases}\right]

⇒ f\left(x\right)f\left(y\right) = \left[\begin{cases}\cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{cases}\right]\left[\begin{cases}\cos y & - \sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{cases}\right]

⇒ f\left(x\right)f\left(y\right) = \left[\begin{cases}\cos x \cos y - \sin x \sin y & - \cos x \sin y - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & - \sin x \sin y + \cos x \cos y & 0 \\ 0 & 0 & 1\end{cases}\right]

⇒ f\left(x\right)f\left(y\right) = \left[\begin{cases}\cos ( x + y ) & - \sin ( x + y ) & 0 \\ \sin ( x + y ) & \cos ( x + y ) & 0 \\ 0 & 0 & 1\end{cases}\right]

$⇒ f ( x ) \cdot f ( y ) = f ( x + y )$

Hence statement-II is also correct.