The action in the Chern-Simons theory is given as:
$$S=\frac{k}{4\pi}\int_M \text{Tr}(A \: \wedge \:dA + \frac{2}{3}A \: \wedge \:A \: \wedge \: A). $$ The wikepida page gives the euler lagrange equations as $0=\frac k{2\pi}F= \frac k{2\pi}dA+A\wedge A$.
Here is my attempt $$ \delta S=\frac{k}{4\pi}\delta\int_M \text{Tr}(A \: \wedge \:dA + \frac{2}{3}A \: \wedge \:A \: \wedge \: A) $$ Assuming $\delta$ commutes with taking the integral and taking the trace $$ =\frac{k}{4\pi}\int_M \delta\text{Tr}(A \: \wedge \:dA + \frac{2}{3}A \: \wedge \:A \: \wedge \: A) $$ $$ =\frac{k}{4\pi}\int_M \text{Tr}(\delta(A \: \wedge \:dA + \frac{2}{3}A \: \wedge \:A \: \wedge \: A)) $$ $$ =\frac{k}{4\pi}\int_M \text{Tr}(\delta(A \: \wedge \:dA + \frac{2}{3}A \: \wedge \:A \: \wedge \: A)) $$
$$ =\frac{k}{4\pi}\int_M \text{Tr}(\delta(A \: \wedge \:dA)) + \text{Tr}(\delta(\frac{2}{3}A \: \wedge \:A \: \wedge \: A)) $$ $\delta$ has a Leibniz rule $$\text{Tr}(\delta(A \: \wedge \:dA))=\text{Tr}(\delta A \: \wedge \:dA+ A \: \wedge \:\delta dA)$$
$$\text{Tr}(\delta(\frac{2}{3}A \: \wedge \:A \: \wedge \: A))=\text{Tr}(\frac{2}{3}(\delta A \: \wedge \:A \: \wedge \: A+A \: \wedge \:\delta A \: \wedge \: A+A \: \wedge \:A \: \wedge \: \delta A))=\text{Tr}(\frac{2}{3}(A \: \wedge \:A \: \wedge \: \delta A))$$
Putting this back together: $$ \delta S=\frac{k}{4\pi}\delta\int_M \text{Tr}(\delta A \: \wedge \:dA+ A \: \wedge \:\delta dA) +\text{Tr}(\frac{2}{3}(A \: \wedge \:A \: \wedge \: \delta A)).$$
I do not see how to continue this further. This answer attempts this question another way. I do understand how they conclude from $ 0=\frac{k}{4 \pi} \int_M \text{Tr} \bigg[a \: \wedge \: (2dA \: + 2A \: \wedge \: A) \bigg]$ that $\frac{k}{2 \pi}F=0.$