jmc

We can factor the numerator and denominator to get $\frac{x^3-x^2+x}{6x^2-9x}=\frac{x(x^2-x+1)}{3x(2x-3)}.$ In this representation we can immediately see that there is a hole at $x=0$, and a vertical asymptote at $x=\frac{3}{2}$. There are no more holes or vertical asymptotes so $a=1$ and $b=1$. If we cancel out the common factors, our rational function simplifies to $$\frac{x^2-x+1}{3(2x-3)}.$$We see that as $x$ becomes very large, the $x^2$ term in the numerator will dominate. To be more precise, we can use polynomial division to write $\frac{x^2-x+1}{3(2x-3)}$ as $$\frac{2x+1}{12}+\frac{7}{12(2x-3)},$$from which we can see that for large $x,$ the graph tends towards $\frac{2x+1}{12},$ giving us an oblique asymptote.

Since the graph cannot have more than one oblique asymptote, or an oblique asymptote and a horizontal asymptote, we have that $c=0$ and $d=1$. Therefore, $a+2b+3c+4d=1+2+0+4=\boxed{7}.$