Manipulating the function:
[tex]f(x)=\dfrac{2e^x+3e^{3x}}{e^{2x}-e^{3x}}=\dfrac{2+3e^{2x}}{e^{x}-e^{2x}}.~\text{Let}~e^x=u.~\text{If}~x\to\infty,~\text{so}~u\to\infty.\\\\
\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\dfrac{2+3e^{2x}}{e^{x}-e^{2x}}=\lim_{u\to\infty}\dfrac{2+3u^2}{u-u^2}=\\\\
\lim_{u\to\infty}\dfrac{u^2(\frac{2}{u^2}+3)}{u^2(\frac{1}{u}-1)}=\lim_{u\to\infty}\dfrac{\frac{2}{u^2}+3}{\frac{1}{u}-1}=\dfrac{3}{-1}=-3.\\\\
\text{So, the assymptote of f(x) is:}\\\\
\boxed{\lim_{x\to\infty}f(x)=-3}\Longrightarrow\boxed{\boxed{y=-3}}[/tex]
