Find an orthonormal basis for $R_2[x]$ where $\langle f(x),g(x)\rangle=\int_0^1 f(x)g(x)\,dx$
I started with $\{1,x,x^2\}$
$$\|v_1\|=\sqrt{\langle v_1,v_1\rangle}=\int_0^1\ dx=\sqrt{1}=1$$
So $e_1=1$
$$v'_2=v_2-\langle v_2,e_1\rangle e_1=x-\int_0^1x\ dx=x-\frac{1}{2}$$
$$\|v_2\|=\sqrt{\langle v_2,v_2\rangle}=\sqrt{\int_0^1 (x-\frac{1}{2})^2\ dx}=\frac{1}{\sqrt{12}}$$
$$e_2=\sqrt{12}(x-\frac{1}{2})$$
\begin{align} v'_3=v_3-\langle v_3,e_1\rangle e_1-\langle v_3,e_2\rangle e_2 & =x^2-\int_0^1 x^2\ dx-12\int_0^1(x^4-\frac{x^3}{2}\ dx)(x-\frac{1}{2}) \\[10pt] &=x^2-\frac{9}{10}x+\frac{7}{60} \end{align}
which seems to be wrong as $e_3=\sqrt{180}(x^2-x+\frac{1}{6})$
Your functions $e_1$ and $e_2$ are correct, but you made a mistake in the computation of $v_3'(x)$. You should have obtained$$v_3'(x)=x^2-x+\frac16.$$And, since $\|v_3'\|=\frac1{6\sqrt5}$, $e_3(x)=6\sqrt5\left(x^2-x+\frac16\right)$.