Find the sum of the inverse roots of $x^3-7x^2+4x-1$

If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-\frac{4}{-1}=4$.

Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$\frac1{x_1}+\frac1{x_2}+\frac1{x_3}=\frac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=\frac{-c/a}{d/a}=-\frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.