find $g$ such that $g\circ f=h$ Let 
$$f(x)=\dfrac{2x+3}{x-1} \quad \mbox{and}\quad h(x)=\dfrac{6x^2+8x+11}{(x-1)^2} $$
find polynom $g$ such that $g\circ f=h$
Indeed,
$$g\circ f=h \iff \forall x\in \mathbb{R}\quad g\circ f(x)=h(x) $$
\begin{align}
g\circ f(x)&=h(x)\\
g\left(f(x)\right)&=h(x)\\
g\left(\dfrac{2x+3}{x-1}\right)&=\dfrac{6x^2+8x+11}{(x-1)^2}
\end{align}
Let $y=\dfrac{2x+3}{x-1}$
 \begin{align}
y&=\dfrac{2x+3}{x-1}\\
x&=\dfrac{3+y}{y-2}
\end{align}
\begin{align}
g\left(y\right)&=\dfrac{6\left(\dfrac{3+y}{y-2}\right)^2+8\left(\dfrac{3+y}{y-2}\right)+11}{\left(\dfrac{3+y}{y-2}-1\right)^2}\\
&=\dfrac{6\left(\dfrac{3+y}{y-2}\right)^2+8\left(\dfrac{3+y}{y-2}\right)+11}{\left(\dfrac{3+y}{y-2}-1\right)^2} \\
&=\dfrac{6\left(3+y\right)^2+8\left(3+y\right)+11\left(y-2 \right)}{25} \\
&=\dfrac{6y^2+55y+56}{25} \\
\end{align}
Then  $g(x)=\dfrac{6x^2+55x+56}{25}$ such that $g\circ f=h$
AM i right ?

It is wrong. After you have written the expression of $g(y)$ we should have
$$ g(y) = \frac{6(3+y)^{2} + 8(3+y)(y-2) + 11(y-2)^{2}}{25} $$
instead of your expression. On simplifying we get $g(y) = y^{2}+2$, i.e,

$$ g(x) = x^{2} + 2$$

