help finding the PDF of this function

Question

the question I'm trying to solve is let x be a continuous uniform random variable in the interval [0,3] with $$ y= 2x-x^2 $$ what I did so far is used the method of transformation, and solved for x using the quadratic formula getting, $$ 1\pm \sqrt{1-y} $$ then finding the derivative I got $$ \pm \frac{1}{2\sqrt{1-y}} $$ combining everything together with the method of transformation formula: $$ f_x(f^{-1}(Y)) \frac{dh^{-1}}{dy} = \frac{1}{3}\left(\pm\frac{1}{2\sqrt{1-y}}\right)= \pm\frac{1}{6\sqrt{1-y}} ~\text{for}~ -3<y<0 $$ i got the bounds by subsituing x with 0 and 3, but when i try proving this is a PDF by finding the integral from -3 to 0 i dont get 1, i get 1/3. but i realize if i dont multiply 1/3 and just find the integral of the inverse function i do get 1 i.e the integral $$ \frac{1}{2\sqrt{1-y}} $$ from -3 to 0 the answer is 1. any help is appreciated