9. 3. Let $f(x)$ be a quadratic trinomial. The trinomial $x^{2}$ - $f\left(\frac{1}{x}+1\right)$ or $(x-1)^{2} f\left(\frac{1}{x}-1\right)$ can replace $f(x)$. Can these operations transform the quadratic trinomial $x^{2}+4 x+3$ into the quadratic trinomial $x^{2}+10 x+3$?
9.3. Not possible.

Let $f(x)=A x^{2}+B x+C$, then $\Delta=B^{2}-4 A C$. After performing the first operation, $f(x)$ becomes
$$
\begin{array}{l}
(A+B+C) x^{2}+(B+2 A) x+A, \\
\Delta=(B+2 A)^{2}-4(A+B+C) A=B^{2}-4 A C .
\end{array}
$$

While performing the second operation, we get
$$
\begin{array}{l}
C x^{2}+(B-2 C) x+(A-B+C), \\
\Delta=(B-2 C)^{2}-4 C(A-B+C)=B^{2}-4 A C .
\end{array}
$$

Therefore, after performing the two specified operations, the discriminant remains unchanged. However, the discriminant of $x^{2}+4 x+3$ is 4, while the discriminant of $x^{2}+10 x+9$ is 64. Therefore, it is not possible to obtain the quadratic trinomial $x^{2}+10 x+9$ from the quadratic trinomial $x^{2}+4 x+3$.