Find the derivative of the function $f(x)=|x^3-4x^2+3x-2|$ and solve for $x$ when $f^{\prime }(x)=0$

Step 1 ：Step 1: Identify the inside function of the absolute value as $g(x)=x^3-4x^2+3x-2$.

Step 2 ：Step 2: Find the derivative of $g(x)$, $g^{\prime }(x)=3x^2-8x+3$.

Step 3 ：Step 3: The derivative of the absolute value function is $f^{\prime }(x)=g^{\prime }(x)$ when $g(x)>0$ and $f^{\prime }(x)=-g^{\prime }(x)$ when $g(x)<0$. So we need to find the values of $x$ when $g(x)=0$.

Step 4 ：Step 4: Solve for $x$ in the equation $g(x)=0$, which gives $x$ approximately equal to 0.554, 1.223 and 2.723.

Step 5 ：Step 5: Check the sign of $g^{\prime }(x)$ in intervals determined by the roots of $g(x)$. We find that $g^{\prime }(x)>0$ when $x<0.554$ or $1.223<x<2.723$, and $g^{\prime }(x)<0$ when $0.554<x<1.223$ or $x>2.723$.

Step 6 ：Step 6: Therefore, $f^{\prime }(x)=g^{\prime }(x)$ when $x<0.554$ or $1.223<x<2.723$, and $f^{\prime }(x)=-g^{\prime }(x)$ when $0.554<x<1.223$ or $x>2.723$.

Step 7 ：Step 7: Set $f^{\prime }(x)=0$ and solve for $x$. We find that $x$ approximately equal to 0.333 and 2.667.