Suppose $V$ is a finite-dimensional vector space and let $v\in V$ and $\varphi\in V^*$. Find the image of operator which corresponds to the tensor $v\otimes \varphi$.
My approach: Since $V\cong V^{**}$ in canonical way then we can consider vector $v$ as a tensor of type $(0,1)$. And $\varphi$ is tensor of type $(1,0)$. Then $v\otimes \varphi\in T_1^1(V)$, where by $T_p^q(V)$ I mean the linear space of all tensors of type $(p,q)$.
Also we know that there is canonical isomorphism between $L(V)$ and $T_1^1(V)$, where by $L(V)$ I denoted the linear space of all operators on $V$. This canonical isomorphism is given by $\pi:L(V)\to T_1^1(V)$ such that $f\mapsto \pi_f$ s.t. $\pi_f(u,\xi)=\xi(f(u))$ for all $u\in V, \xi\in V^*$.
By above remark it follows that there is an operator $f$ on $V$ s.t. $\pi_f=v\otimes \varphi$. It means that for all $u\in V, \xi\in V^*$ we have $\xi(f(u))=\xi(v)\varphi(u)$.
Let $y\in \operatorname {Im}f$ then $y=f(x),x\in V$. Then $\xi(f(x))=\xi(v)\varphi(x)$ and let $\alpha:=\varphi(x)$ then $\xi(f(x)-\alpha v)=0$ for all $\xi in V^*$ then it means that $f(x)=\alpha v$ i.e. $y=\alpha v\in \langle v\rangle $, i.e. $\operatorname {Im}f\subseteq \langle v\rangle$.
However the converse inclusion is true if $\varphi \neq 0$ because if $\varphi=0$ then $v\otimes \varphi=0$ and corresponding operator $f$ is zero operator. But the image of such operator is $\{0\}$ which may not be $\langle v\rangle$.
Assume that $\varphi\neq 0$ then $\exists x\in V$ s.t. $\varphi(x)\neq 0$. Then $\xi(f(x))=\varphi(x)\xi(v)$ and let $\lambda=\varphi(x)\neq 0$ and then $\xi(f(x)-\lambda v)=0$ for all $\xi \in V^*$ which implies that $f(x)=\lambda v$ then $v=f(\lambda^{-1}x)$. Hence $v\in \operatorname{Im}f$ then $\langle v\rangle \subseteq \operatorname{Im}f$.
In summary, if $\varphi\neq 0$ then $\operatorname{Im}f=\langle v\rangle$.
Is this reasoning correct? Would be very grateful for any comments, please.
Yes, your reasoning is correct. In fact, note that the inverse of the isomorphism $L(V) \to T_1^{~1}(V)$ you wrote is the unique linear map $T_1^{~1}(V) \to L(V)$ characterized by $$T_1^{~1}(V) \ni v\otimes \varphi \mapsto \big(x\mapsto\varphi(x)v \big)\in L(V), $$and $x \mapsto \varphi(x)v$ has rank $1$ if $\varphi$ and $v$ are non-zero, and it is the zero operator if either $\varphi$ or $v$ is zero.