For the function:
$f(x) = 1-6x^2 + 4x^3$ for $0 \leq x \leq 1$ and requiring that $f(x+2) = f(x)$ and $f(-x) =f(x)$ for all $x \in \mathbb{R}$,
it gives us a hint that is to: consider $f'(0),f'(1)$ and $f'''(x)$.
I found them all, and see no use of them.
So letting the fourier series of $f$ be
$Sf(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty}a_k \cos k\pi x$, (no sine term since the function is even)
then using the hints I acquire the results (and also using the fact that a fourier series will converge since the function is piecewise continuous)
$f'(0) = 0 = \sum_{k=1}^\infty k\pi a_k \sin( k \pi 0) = 0$
$f'(1) = 0 = \sum_{k=1}^\infty k\pi a_k \sin( k \pi 1) = 0$
$f'''(x) = 24 = \sum_{k=1}^\infty k^3 \pi ^3 a_k \sin (k \pi x)$.
which say nothing about the Fourier Series.
Is there a way to utilise this hint (as finding the constants $a_k$ will require in total of 6 integration by parts).
Probably, the hint is simply an help for the integrations, nothing more. The simplifications appear in eqs. (1) and (2) below. The integral with $(1-6x^2+4x^3)$ is not explicitly computed. Only $f'(0)=f'(1)=0$ and $f'''(x)=24$ are used. $$a_k=2\int_0^1f(x)\cos(k\pi x)dx$$ $$a_k=\big[\frac{2}{k\pi}\sin(k\pi x)f(x)\big]_0^1 -\frac{2}{k\pi}\int_0^1f'(x)\sin(k\pi x)dx=-\frac{2}{k\pi}\int_0^1f'(x)\sin(k\pi x)dx$$ $$a_k=\big[\frac{-2}{k^2\pi^2}\cos(k\pi x)f'(x)\big]_0^1 +\frac{2}{k^2\pi^2}\int_0^1f''(x)\cos(k\pi x)dx$$ $$f'(0)=f'(1)=0 \quad\to\quad a_k=\frac{2}{k^2\pi^2}\int_0^1f''(x)\cos(k\pi x)dx \tag 1$$ $$a_k=\big[\frac{2}{k^3\pi^3}\sin(k\pi x)f''(x)\big]_0^1 -\frac{2}{k^3\pi^3}\int_0^1f'''(x)\sin(k\pi x)dx=-\frac{2}{k^3\pi^3}\int_0^1f'''(x)\sin(k\pi x)dx$$ $$f'''(x)=24 \quad\to\quad a_k=-\frac{48}{k^3\pi^3}\int_0^1\sin(k\pi x)dx \tag 2$$ $$a_k=\frac{48(1-(-1)^k)}{k^4\pi^4}$$