Step 1: Find the domain first.
Since $\log(2x-3)$ needs a positive argument, we require $2x-3>0$, that is $x>\dfrac{3}{2}$. So the whole discussion lives on $\left(\dfrac{3}{2},\infty\right)$.
Step 2: Differentiate the function.
For $f(x)=\log(2x-3)-2x^2+6x-4$, term by term we get $f'(x)=\dfrac{2}{2x-3}-4x+6$.
Step 3: Tidy the derivative.
Notice $-4x+6=-2(2x-3)$, so $f'(x)=\dfrac{2}{2x-3}-2(2x-3)=\dfrac{2-(2x-3)^2}{2x-3}$.
Step 4: Set up the increasing condition.
We need $f'(x)>0$. On the domain the denominator $2x-3$ is positive, so the sign is decided by the numerator $2-(2x-3)^2>0$.
Step 5: Solve the numerator inequality.
$(2x-3)^2<2$ means $-\sqrt2<2x-3<\sqrt2$, so $\dfrac{3-\sqrt2}{2}<x<\dfrac{3+\sqrt2}{2}$, roughly $0.79<x<2.21$.
Step 6: Intersect with the domain.
Combining with $x>\dfrac{3}{2}$ gives the increasing interval $\left(\dfrac{3}{2},\dfrac{3+\sqrt2}{2}\right)$, which is essentially $\left(\dfrac{3}{2},2\right)$ as listed.
\[ \boxed{\left(\dfrac{3}{2},\,2\right)} \]