Your derivative is correct...here are the steps...
\(\begin{array}\ y&=&x^{\frac12} \cdot(3-x)^2 \\ \frac{dy}{dx}&=&(\frac{d}{dx} x^{\frac12})(3-x)^2+(\,\frac{d}{dx}(3-x)^2\,)(x^{\frac12}) \\ \frac{dy}{dx}&=&(\frac12)(x^{-\frac12})(3-x)^2+(2)(3-x)(\,\frac{d}{dx}(3-x)\,)(x^{\frac12}) \\ \frac{dy}{dx}&=&(\frac12)(x^{-\frac12})(3-x)^2+(2)(3-x)(-1)(x^{\frac12}) \\ \frac{dy}{dx}&=&(3-x)(\quad(\frac12)(x^{-\frac12})(3-x)+(2)(-1)(x^{\frac12})\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(\quad(x^{-\frac12})(3-x)-4x^{\frac12}\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(x^{-\frac12})(\quad(3-x)-4x\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(x^{-\frac12})(3-5x) \end{array}\)
The above equation tells us what the slope of this curve is at any x .
Where the tangent line is horizontal, the slope of the curve = 0 .
Where the tangent line is horizontal, the dy/dx of the curve = 0 .
We want to know what values for x cause dy/dx to be 0 .
\(0\,=\,(3-x)(\frac12)(x^{-\frac12})(3-5x) \)
Set the first and fourth factors equal to zero and solve for x .
0 = 3 - x 0 = 3 - 5x
x = 3 5x = 3
x = 3/5
You're right that the minumum is a point where the tangent line is horizontal...but the maximum is also a point where the tangent line is horizontal.