Fourier series coefficient of signal when Time period is twice the fundamental period
As Matt stated, by your calculation you have excluded options a and b but you should also check for options c and d, from which you would see that the answer is the option d.

You could practically check the result of coefficients $d_k$ from the interpretation that CTFS coefficients of a periodic waveform $\tilde{x}(t)$ can be obtained from an inverse-DTFT (or an inverse-DFT more practically), by treating $\tilde{x}(t)$ as a DTFT waveform. Note that the computation is exact if the CTFS coefficients $d_k$ were of finite length in advance, but otherwise approximate which would only get better by choosing long enough samples to represent the waveform $\tilde{x}(t)$.

Therefore, the following computes (approximately here) $3N$ CTFS coefficients of $x(t)$: 

&gt;     N = 128;
&gt;     x = [ones(1,N) , -1*ones(1,N) , zeros(1,N)];
&gt;     d = ifft(x);


On the orther hand the following code would do the same by treating $\tilde{x}(t)$ as periodic in $2 T_0$ rather than $T_0$

&gt;     N = 128;
&gt;     x = [ones(1,N) , -1*ones(1,N) , zeros(1,N)];
&gt;     d = ifft([x x]);


Based on your edit, the following provides you the answer:
Given that the CTFS coefficients of the derivative signal is:

\begin{align}
d_n &amp;=\frac{1}{6} \int_{-3}^3 [\delta(t+3)-2\delta(t+2)+\delta(t+1)+\delta(t)-2\delta(t-1)+\delta(t-2)]e^{\frac{-jn2\pi t}{6}}dt\\
&amp;=\frac{1}{6}[  e^{j\frac{2\pi}{6}3n} - 2 e^{j\frac{2\pi}{6}2n} + e^{j\frac{2\pi}{6}1n} + 1 -2e^{-j\frac{2\pi}{6}1n} + e^{-j\frac{2\pi}{6}2n}] \\
&amp;=\frac{1}{6} [  e^{j\pi n} - 2 e^{j\frac{2\pi}{3}n} + e^{j\frac{\pi}{3}n} + 1 -2e^{-j\frac{\pi}{3}n} + e^{-j\frac{2\pi}{3}n}] \\
\end{align}

Above three lines were straightforward. Now we shall group those terms to yield something that can be simplified:

$$d_n =\frac{1}{6} [  (1 + e^{j\pi n}) - 2 ( e^{j\frac{2\pi}{3}n} + e^{-j\frac{\pi}{3}n} ) + (e^{j\frac{\pi}{3}n} + e^{-j\frac{2\pi}{3}n})] $$

Now add $2\pi n$ to the negative angles so that they become: $$e^{-j\frac{\pi}{3}n} = e^{j( 2\pi n -\frac{\pi}{3}n) } = e^{j\frac{5\pi}{3}n } = e^{j\pi n }e^{j\frac{2\pi}{3}n } $$ and  $$e^{-j\frac{2\pi}{3}n} = e^{j( 2\pi n -\frac{2\pi}{3}n) } = e^{j\frac{4\pi}{3}n } = e^{j\pi n}e^{j\frac{\pi}{3}n } $$ respectively. And plugging them into $d_n$ line, yields:

\begin{align}
d_n &amp;=\frac{1}{6} [  (1 + e^{j\pi n}) - 2 ( e^{j\frac{2\pi}{3}n} + e^{j\pi n }e^{j\frac{2\pi}{3}n } ) + (e^{j\frac{\pi}{3}n} + e^{j\pi n}e^{j\frac{\pi}{3}n})]\\ 
&amp;=\frac{1}{6} [  (1 + e^{j\pi n}) - 2 (1 + e^{j\pi n}) e^{j\frac{2\pi}{3}n} + (1 + e^{j\pi n}) e^{j\frac{\pi}{3}n}]\\ 
&amp;=\frac{1}{6} [1 + e^{j\pi n}]\cdot[1 - 2 e^{j\frac{2\pi}{3}n} + e^{j\frac{\pi}{3}n}]\\
&amp;=\frac{1}{6} [e_n]\cdot[f_n]\\
\end{align}

Now it can be shown that the product term $e_n$ will be zero for all odd indice $n$, hence $d_n$ will also be zero for all odd $n$. Note that the other product term $f_k$ will be zero for $n = 6m$ that $d_n$ will also be zero for $n = 6,12,18...$ However for your example it sufficies to show that $e_n$ will be zero for odd $n$.

FURTHERMORE below is the theoretical proof that it's not a coincidence to have all the odd indexed terms in CTFS to be zer0, independent of the signal itself, when the period $T_y$ of the periodic signal $x(t)$ is assumed to be twice that of its fundamental period $T_x$. Assume $x(t)$ is a periodic signal with fundamental period of $T_x$ and let $y(t)$ be the same signal $x(t)$ but interpreted to have a period of $T_y = 2 T_x$. Let $a_k$  and $b_k$ denoted the CTFS coefficients of $x(t)$ and $y(t)$ respectively, then we have:

\begin{align}
b_k &amp;= \frac{1}{T_y} \int_{0}^{T_y} y(t) e^{-j\frac{2\pi}{T_y} k t} dt = \frac{0.5}{T_x} \int_{0}^{2T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt\\
&amp;= 0.5 \left( \frac{1}{T_x} \int_{0}^{T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt + \frac{1}{T_x} \int_{T_x}^{2T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt \right)\\
\end{align}

Make the substitution $t' = t - T_x$ in the second integral and recognize that $x(t'+T_x) = x(t')$ as $x(t)$ is periodic with $T_x$ , yielding:

$$b_k = 0.5 \left( \frac{1}{T_x} \int_{0}^{T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt + e^{-j\frac{2\pi}{T_x} (k/2) T_x} \frac{1}{T_x} \int_{0}^{T_x} x(t') e^{-j\frac{2\pi}{T_x} (k/2) t'} dt' \right) $$

Now recognize the integrals as $a_{k/2}$; i.e, the CTFS of the signal $x(t)$ evaluated at $k/2$. And simplify the sum:
\begin{align}
b_k &amp;= 0.5 \left( a_{k/2}+ e^{j\pi k} a_{k/2} \right)\\
&amp;= 0.5 \left( 1+ e^{j\pi k} \right) a_{k/2} \\
&amp;= \begin{cases}{ a_{k/2} ~~, \text{ for k even, k=2m  , m=1,2,...}\\ 0 ~~~~~~~, \text{ for k odd  , k=2m+1, m=1,2,...}}\end{cases} \\
\end{align}