Find the range of $ f(x)=9^x - 3^x+1$ 
Problem:-Find range of function $ f(x)=9^x - 3^x+1$, here the domain of $f$ is $\mathbb R$.
Solution: $ f(x)=9^x - 3^x+1$. Let $f(x)=y$. Then 
  $$ \begin{split}y&=9^x - 3^x+1\\
 y&=3^{2x} - 3^x+1
\end{split}$$
  Let $3^x= u$. Then $ y=u^2 - u+1$, so
  $$ u^2 - u+1-y=0.$$

Am I doing right?

Looks good. To finish it off, we complete the square and notice that:
$$
y = u^2 - u + 1 = \left(u^2-u+\dfrac{1}{4}\right)+\dfrac{3}{4} = \left(u-\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge 0+\dfrac{3}{4}=\dfrac{3}{4}
$$
since the square of any real number must be nonnegative. Hence, the range is:
$$
\{y\in \mathbb{R} \mid y \ge3/4\}
$$
