Sum of the coefficients of $x^3$ and $x^6$ in the expansion of  ${\left(x^2-\frac{1}{x}\right)}^9$is 

${\mathrm{T}}_{r+1}{=}^9{C}_r{\left(x^2\right)}^{9-r}{\left(-\frac{1}{x}\right)}^r=(-1{)}^r\left({ }^9{C}_r\right)x^{18-3r}$

For coefficient of  $x^3,x^6\text{,}$ we set $18-3r=3,6$

$\Rightarrow r=5,4$

$\therefore$  sum of coefficients of  $x^3$ and $x^6$ 

$=(-1{)}^5\left({ }^9{C}_4\right)+(-1{)}^4\left({ }^9{C}_5\right)=0$