Finding $f'(a)$ for a given value

$f(a)$ is equal to $f(a) = 3a^2 - 4a + 1$. The $a$ is left indicated, because it is only given "indicated" to you, not as an actual value.

The derivative is equal to the limit: $$\begin{align*} f'(a) &= \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\\ &= \lim_{h\to 0}\frac{ \bigl(3(a+h)^2 -4(a+h) + 1\bigr) - \bigl(3a^2-4a+1\bigr)}{h}\\ &=\lim_{h\to 0}\frac{\bigl( 3(a^2+2ah+h^2) - 4a-4h +1\bigr) - \bigl(3a^2-4a+1\bigr)}{h}\\ &=\lim_{h\to 0}\frac{3a^2 + 6ah + 3h^2 - 4a - 4h + 1 - 3a^2 + 4a - 1}{h}\\ &=\lim_{h\to 0}\frac{3h^2 + 6ah - 4h}{h}. \end{align*}$$ Up to here, all we've done is use the definition of $f$, and do algebra on the expression in the limit, nothing else.

At this point: if you try plugging in $h=0$ into the limit, you get $\frac{0}{0}$, which is what you should expect (computing limits of difference quotients always give $\frac{0}{0}$ at first). But there is clearly a factor of $h$ to be factored out of the numerator; factor it, cancel it with the denominator, and do the resulting limit. The answer should be an expression that involves $a$ but not $h$ (since the question is in terms of $a$, the answer will be in terms of $a$ as well).

(The whole point here is to realize that for any particular value of $a$, the computations of the limit, and so of the derivative, are actually the same: substitue $a=1$ and you can do exactly the same steps as above to get the value $f'(1)$; substitute $a=2$, and again the exact same steps work to find $f'(2)$; substitute $a=\pi$, and the same steps work to compute $f'(\pi)$; substitute $a=1058431278903210532.5789432\sqrt{2}$, and the same steps work to compute $$f'(1058431278903210532.5789432\sqrt{2}).$$ So instead of doing all the work each time we need the value of the derivative at a point, we can just do the work once, and get an answer into which we will just need to "plug in" whatever number we need to get the answer.)

Do you know the basic rules for computing the derivative of a function?

Assuming so, for $f(x)=3x^2-4x+1$, you have $f'(x)=3\times 2x-4=6x-4$. Then $f'(a)=6a-4$.

If you have to compute $f'(a)$ from the definition, evaluate

$$f^{\prime }(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}.$$

Added in response to OP's comment.

You have $f(x)=3x^{2}-4x+1$. So $f(a)=3a^{2}-4a+1$ and

$$f(a+h)=3\left( a+h\right) ^{2}-4\left( a+h\right) +1.$$

Thus$^1$ $$\begin{eqnarray*} f^{\prime }(a) &=&\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h} \\ &=&\lim_{h\rightarrow 0}\frac{3\left( a+h\right) ^{2}-4\left( a+h\right) +1-\left( 3a^{2}-4a+1\right) }{h} \\ &=&\lim_{h\rightarrow 0}\frac{6ah+3h^{2}-4h}{h}\quad \text{you have to simplify this fraction as shown below} \\ &=&\lim_{h\rightarrow 0}\; 6a+3h-4 \\ &=&6a-4. \end{eqnarray*}$$

-- $^1$ Detailed computation. From $$\begin{eqnarray*} 3\left( a+h\right) ^{2} &=&3a^{2}+6ah+3h^{2} \\ -4\left( a+h\right) +1 &=&-4a-4h+1 \\ -\left( 3a^{2}-4a+1\right) &=&-3a^{2}+4a-1 \end{eqnarray*}$$

we get $$\begin{eqnarray*} &&3\left( a+h\right) ^{2}-4\left( a+h\right) +1-\left( 3a^{2}-4a+1\right) \\ &=&3a^{2}+6ah+3h^{2}-4a-4h+1-3a^{2}+4a-1=6ah+3h^{2}-4h \end{eqnarray*}$$

and for $h\ne 0$

$$\frac{6ah+3h^{2}-4h}{h}=\frac{h(6a+3h-4)}{h}=6a+3h-4$$