If $D$ be the differentiation operator on $V$. Find $D^*$. 
Let $V$ be the vector space of the polynomials over $R$ of degree less than or
  equal to $3$ with the inner product space $(f|g)=\int_{0} ^{1}f(t)g(t) dt$, and let $D$ be the differentiation operator on V. Find $D^*$

Attempt
As I did the calculation was very large, so I will explain what I did step by step so that you say if I made a mistake or I hit
STEP 1: First I considered a canonical basis of $R _{\leq 2} [X]$ that is $ \{ 1,x,x^2,x^3\}$. I did the Gram-Schmidt Process to orthogonalize this base. Resulted in
$$ \left\{ 1, x-\frac{1}{2}, x^2-x+ \frac{1}{6}, x^3-\frac{3}{2}x^2+\frac{3}{5}x- \frac{1}{2} \right\}$$
STEP 2: 
Orthonormalize this base, dividing each element by its norm (in this part I may have been wrong because they gave many calculations). Resulted in
$$ \{ 1, \sqrt{12} \left(x-\frac{1}{2} \right) , \sqrt{180} \left( x^2-x+ \frac{1}{6} \right) , \sqrt{\frac{33600}{18772}} \left( x^3-\frac{3}{2}x^2+\frac{3}{5}x- \frac{1}{2} \right) \}$$
STEP 3: Write the matrix of $D$
$$\begin{bmatrix}0&\sqrt{12}&0& \frac{1}{10} \sqrt{\frac{33600}{18772}}\\0&0&2\sqrt{15}& 0\\ 0&0&0& \sqrt{\frac{33600}{33378960}} \\ 0&0&0& 0 \end{bmatrix}$$
STEP4: The matrix of $D^*$ is the conjugate transpose of the matrix of $D$,(in this case only transpose) according to the corollary: 

Let $V$ be a finite-dimensional inner product space, and let
  $T$ be a linear operator on $V$. In any orthonormal basis for $V$, the matrix of $T^*$ is the conjugate transpose of the matrix of $T$.

At least the idea is correct? Is there any easier way to do this exercise?

Yes, your idea is correct, and this is the standard way to find the adjoint of an operator.
Another attempt could be using integration by parts:
\begin{align}
\langle f',g\rangle &= \int_0^1 f'(t)g(t)\,dt \\
&= f(t)g(t)\bigg|_0^1 - \int_0^1 f(t)g'(t)\,dt \\
&= f(1)g(1) - f(0)g(0) - \langle f,g'\rangle \\
\end{align}
Hence find an operator $A$ such that $\langle f,Ag\rangle =  f(1)g(1) - f(0)g(0)$ and then we will have $$\langle f',g\rangle = \langle f,(A-D)g\rangle$$
Therefore the adjoint is $D^* = A-D$.

To find $A$ explicitly, let's find a polynomial $p(x) = ax^3+bx^2+cx+d \in \mathbb{R}_{\le 3}[x]$ such that for all $f \in \mathbb{R}_{\le 3}[x]$ holds
$$\int_0^1 f(t)p(t)\,dt = f(0)$$
In particular, for $f(x) = 1,x,x^2,x^3$ we get
$$1 = \int_0^1 p(t)\,dt = \frac{a}4 + \frac{b}3 + \frac{c}2 + d$$
$$0 = \int_0^1 tp(t)\,dt = \frac{a}5 + \frac{b}4 + \frac{c}3 + \frac{d}2$$
$$0 = \int_0^1 t^2p(t)\,dt = \frac{a}6 + \frac{b}5 + \frac{c}4 + \frac{d}3$$
$$0 = \int_0^1 t^3p(t)\,dt = \frac{a}7 + \frac{b}6 + \frac{c}5 + \frac{d}4$$
Solving the system yields $p(x) = -140x^3+ 240x^2 -120x +16$.
Similarly we find that the polynomial $q(x) = 140x^3-180x^2 +60x -4$ satisfiels
$$\int_0^1 f(t)q(t)\,dt = f(1), \quad\forall f \in \mathbb{R}_{\le 3}[x]$$
Therefore, if we set $Ag := (p-q)g$, we get
$$\langle f,Ag\rangle = \int_0^1f(t)g(t)(p(t) - q(t))\,dt = f(1)g(1) - f(0)g(0)$$
