Answer
$x=1$ , $x=3$.
Work Step by Step
The slope of the tangent at x=a
is the derivative of f(x), calculated in x=a.
The tangent line has slope $f^{\prime}(a)$, and passes through $(a,f(a)).$
\begin{align*}
f(x)&=(x^{3}-6x^{2}+9x+1)^{1/2} \quad \\&\text{.. apply chain rule...}& \\
f^{\prime}(x)& =\displaystyle \frac{1}{2}(x^{3}-6x^{2}+9x+1)^{-1/2}\cdot(x^{3}-6x^{2}+9x+1)^\prime & \\
& =\displaystyle \frac{1}{2}(x^{3}-6x^{2}+9x+1)^{-1/2}\cdot(3x^{2}-12x+9) & \\
& =\displaystyle \frac{3(x^{2}-4x+3)}{2\sqrt{x^{3}-6x^{2}+9x+1}}\quad & \\
\end{align*}
The derivative is zero when
\begin{align*}
3(x^{2}-4x+3)&=0 \quad & \\
x^{2}-4x+3&=0 \quad & \\
& \text{ ...factoring, } & \\
(x-1)(x-3)&=0 \quad & \\
x&=1, \quad x=3
\end{align*}
The tangent line is horizontal at $x=1$ and $x=3$.