Suppose that $f: G \rightarrow H$ is a homomorphism, and $K$ is a subgroup of $H$. Then the inverse image of $K$ under $f$ is denoted by $f^{-1}(K) = \{x \in G : f(x) \in K\}$. I am asked to show that $f^{-1}(K)$ is a subgroup of $G$, where $f^{-1}(K)$ is a subset of $G$ and not some inverse function if f is a bijection.
What I've tried is: Let $x,y \in f^{−1}(k)$. By definition, then $f(x),f(y) \in K$. Since $K$ is a subgroup of $H$,we have $f(x)f(y)^{−1} \in K$. Since $f$ is a homomorphism, $f(x)f(y)^{−1} = f(xy^{−1})$. So from $f(xy^{−1}) \in K \implies xy^{−1} \in f^{−1}(K)$. Let $e, e′$ be the identities of $G$ and $H$ respectively, then $e′ = f(e) \in K$. So, $e \in f^{−1}(K)$. Thus $f^{−1}(K)$ is a subgroup of $G$.
is my proof sufficient?
Your proof seems fine, although it would benefit from:
Stating explicitly that $f^{-1}(K)\subseteq G$ and
Specifying that you are using the one-step subgroup test. (Some authors define a subgroup $S$ of a group $T$ to be a nonempty subset of $T$ such that for all $r,s\in S$, we have $rs^{-1}\in S$, which could be where you are coming from, but most authors don't.)