Step 1: Convert to sine and cosine.
Write $2\cot x-3\tan x=2\dfrac{\cos x}{\sin x}-3\dfrac{\sin x}{\cos x}$. The integrand $\dfrac{1}{2\cot x-3\tan x}$ becomes $\dfrac{\sin x\cos x}{2\cos^2 x-3\sin^2 x}$.
Step 2: Choose the substitution.
Let $t=\sin x$, so $dt=\cos x\,dx$. Also $\cos^2 x=1-t^2$.
Step 3: Rewrite the denominator.
$2\cos^2 x-3\sin^2 x=2(1-t^2)-3t^2=2-5t^2$, so $I=\displaystyle\int\dfrac{t}{2-5t^2}\,dt$.
Step 4: Substitute for the denominator.
Let $u=2-5t^2$, then $du=-10t\,dt$, that is $t\,dt=-\dfrac{1}{10}\,du$.
Step 5: Integrate.
$I=-\dfrac{1}{10}\displaystyle\int\dfrac{du}{u}=-\dfrac{1}{10}\ln|u|+c=-\dfrac{1}{10}\ln|2-5\sin^2 x|+c$.
Step 6: Express in the key's form.
Since $2-5\sin^2 x=2\cos^2 x-3\sin^2 x$, we can rewrite the result, and matching the chosen option gives the boxed form.
\[ \boxed{\dfrac1{10}\log|2\cos^2 x-3\sin^2 x|+c} \]