Take 22d): \( \boxed{y = -3 f(-3(x-3))+1 } \). Suppose that \( (x,y)\) lies on \(y = f(x)\).
First deal with the inner: Replacing \(x\) with \(x-3\) translates to the right by 3, so we now have the point \( (x+3, y)\)
Replacing \(x-3\) with \(-3(x-3)\) reflects about the \(y\)-axis, and introduces a dilation factor of \( \frac13\), so we now have the point \( \left( -\frac13x + 3, y\right)\).
Now deal with the outer: Replacing \( f(-3(x-3)\) with -3f(-3(x-3))\) reflects about the \(x\)-axis, and introduces a dilation factor of \(3\), so we now have the point \( \left( -\frac13 x+3, -3y\right) \)
Replacing \( -3f(-3(x-3)) \) with \( -3f(-3(x-3)) \) translates upwards by 1, so we now have the point \( \left( - \frac13x+3, -3y+1\right) \).
So our final point is
\[ \left( - \frac13x+3, -3y+1\right). \]
If you need help with the other one, please post any relevant progress.