Prove the sequence $0 \to A \xrightarrow{\alpha} B \xrightarrow{\beta} C \to 0$ is exact if the sequence $0 \to \hom_R(C,N) \xrightarrow{\bar \beta} \hom_R(B,N) \xrightarrow{\bar \alpha} \hom_R(A,N) \to 0$ is exact.
I have figured out all parts except for showing $ker(\beta) \subset im(\alpha)$. Someone suggested that I consider when $N = B/im(\alpha)$ so I played around with it for a bit but I was not able to get the result I wanted.
On
The following is my consideration, correct me if there are mistakes: Since you have just need to show $ker(\beta) \subset im(\alpha)$, assume that $im(\alpha)$ is contained in $ker(\beta)$ and $im(\alpha) \not = ker(\beta)$, then we have two exact sequence: $$0 \rightarrow A \stackrel{\alpha}{\longrightarrow}ker(\beta) \rightarrow ker(\beta)/im(\alpha) \rightarrow0$$ $$0 \rightarrow ker(\beta) \rightarrow B \stackrel{\beta}{\longrightarrow} C \rightarrow 0$$.Since $Hom(-,I)$ is contravariant exact functor for injective modules $I$, then we get that $$(1) 0 \rightarrow Hom(ker(\beta)/im(\alpha),I) \rightarrow Hom(ker(\beta),I) \rightarrow Hom(A,I) \rightarrow0$$ $$(2)0 \rightarrow Hom(C,I) \rightarrow Hom(B,I) \rightarrow Hom(ker(\beta),I)\rightarrow 0$$ for any injective modules $I$. Now take $I$ to be the injective envelope of $ker(\beta)/im(\alpha)$, so the second term of $(1)$ is not zero. Hence $Hom(ker(\beta),I) \not\cong Hom(A,I)$, then by (2), we get a contradiction.
To show the remaining part $\ker \beta\subseteq \operatorname{im} \alpha$, let $N=B/ \operatorname{im}\alpha$, and $f\colon B\to N$ be the canonical projection. Then $\bar{\alpha} f=f\alpha=0$, so $f\in\ker \bar{\alpha}=\operatorname{im}\bar{\beta}$, so $f=\bar{\beta} g=g\beta$ for some $g$. Then $\ker \beta\subseteq\ker f=\operatorname{im}\alpha$.