Let $A,B,C$ be $G$-modules and $0 \rightarrow A\xrightarrow{f} B\xrightarrow{g} C\rightarrow 0$ be a short exact sequence. Let $X^G$ denote the subset of $X$ fixed by $G$. Then the sequence $$0 \rightarrow A^G \xrightarrow{f_1} B^G\xrightarrow{g_1}C^G$$ is exact (where $f_1,g_1$ are restrictions of $f,g$), but adding $0$ to the right may not give exactness. Let $H^1(G,A)$ denote the first cohomology group of $G$ over $A$; similarly define $H^1(G,B)$, $H^1(G,C)$ .
For $c\in C^G$, choose $b\in B$ such that $g(b)=c$. Then define a derivation (or crossed hom.) $f:G\rightarrow A$ by $f(x)=b^x-b$ (here $b^x=$ action of $x\in G$ on $b\in B$.) That $b^x-b$ is in $A$ follows from the fact that $c$ is in $C^G$; we will not write its proof.
If $b'$ is another preimage of $c$ under $g$, then the corresponding derivation $f':G\rightarrow A$, $f'(x)=b'^x-b'$ differs from $f$ by a principal derivation. This allows to define a function $\delta$ (connecting homomorphism) from $C^G$ to $H^1(G,A)$ by $$\delta(c)=\{f:x\mapsto b^x-b\}.$$ (In RHS, it is not a set, but $\delta(c)$ is the map $f$ given as in bracket.)
Consider the sequence $$C^G \xrightarrow {\delta} H^1(G,A) \xrightarrow{f^*} H^1(G,B)\xrightarrow{g^*} H^1(G,C).$$ Aim is to prove exactness of this sequence. I troubled in exactness at $H^1(G,A)$.
My attempt: Consider $\varphi\in H^1(G,A)$ such that $f^*(\varphi)$ is zero in $H^1(G,B)$ i.e. it is principal derivation from $G$ to $B$. Then there exists $b\in B$ such that $$f^*(\varphi)(x)=b^x-b \mbox{ i.e. } f(\varphi(x))=b^x-b.$$ Apply $g$ and noting that $gf=0$, we get $0=g(b)^x-g(b)$. This is true for all $x\in G$. Thus, $g(b)=c$ (say) is a fixed point of $C$ by $G$.
By definition of $\delta$, for this $c=g(b)$, the image of $c$ under $\delta$ is $$\delta(c)=\mbox{the map } (x\mapsto b^x-b) \mbox{ from $G$ to $A$}.$$ Thus $\delta(c)=f^*(\varphi).$ What is troubling me is following: to prove exactness at $H^1(G,A)$, we take $\varphi\in H^1(G,A)$ such that $f^*(\varphi)$ is zero in $H^1(G,B)$, and we have to show that $\varphi=\delta(c)$ for some $c\in C^G$.
But what I obtained above is $f^*(\varphi)=\delta(c)$. What is wrong in my attempt? Any hint?
I think you're confused by the fact that $b\in B$.
The map $x\mapsto b^x-b$ is a principal derivation from $G$ to $B$, but it takes values in the submodule $A$ of $B$, so you can also regard it as a derivation (not usually principal) from $G$ to $A$. So it represents an element (usually non-zero) of $H^1(G,A)$ when regarded as a derivation to $A$, but represents the zero element of $H^1(G,B)$ when regarded as a derivation to $B$.
It doesn't make sense that $f^\ast(\varphi)=\delta(c)$, since $f^\ast(\varphi)\in H^1(G,B)$ but $\delta(c)\in H^1(G,A)$.
In fact, you have proved that $\delta(c)=\varphi$, as you wanted.