It is a little unclear to me exactly what is denoted by $y^\ast$. This being said, I assume until further notice that
$0 \ne x, y \in \Bbb C^n \tag 1$
are column vectors, and that the notation $y^\ast$ denotes a transposed form of $y$, so that $y^\ast$ is a $1 \times n$ row vector. For example, we might have
$y^\ast = y^T, \tag 2$
the transpose of $y$, or perhaps
$y^\ast = \overline{y^T}, \tag 3$
the conjugate transpose or Hermitian adjoint of $y$. It really doesn't matter which of these options we interpret $y^\ast$ to be, the argument is the essentially the same in either case. But we do need to affirm that $y^\ast$ is a $1 \times n$ row vector, for we are given that
$xy^\ast \in \Bbb C^{n \times n}, \tag 4$
that is, $xy^\ast$ is a square matrix in size $n$.
We may find the range of $xy^\ast$ as follows: for any column vector
$z \in \Bbb C^n \tag 5$
we have
$(xy^\ast)z = x(y^\ast z); \tag 6$
now
$y^\ast z \in \Bbb C, \tag 7$
whence
$x(y^\ast z) = (y^\ast z)x \tag 8$
is a scalar multiple of the vector $x$; that is,
$x(y^\ast z) \in \; \text{span}\{x\} \subset \Bbb C^n, \tag 9$
the one-dimensional subspace generated by $x$. By re-scaling $z$ if necessary, we can force $y^\ast z$ to take any value in $\Bbb C$, provided $y \ne 0$; thus in fact
$\text{range}(y^\ast z) = \text{span}\{x\}, \tag{10}$
which is one-dimensional unless $x = 0$ or $y = 0$, in which case it is simply $\{0\}$.
Note Added Saturday 18 July 2020 5:20 PM PST: On the rank of $xy^\ast$: Upon reading the comments made to this answer by our OP Geux, I would like to point out that the above argument effectively solves the rank question as well; indeed, for any square matrix $A$,
$\text{rank}(A) = \dim \text{range}(A); \tag{11}$
this follows from the observation that each column of $A$ maps to a vector in
$\text{range}(A)$ under the action of right multiplication by a column vector of the form
$e_i = (\delta_{ij}), \; 1 \le j \le n, \tag{12}$
that is, $e_i$ is an $n \times 1$ matrix whose $i$-th component is $1$ and which has $0$ for all other entries. If we write $A$ in columnar form,
$A = [A_1 \; A_2 \; \ldots \; A_n], \tag{13}$
then it is easy to see that
$Ae_i = A_i; \tag{14}$
as $1 \to i \to n$, every $A_i$ is thus mapped into $\text{range}(A)$, and it is clear that arbitrary column vectors map to linear combinations of the columns of $A$, which shows that $\text{range}(A)$ is in fact the span of the column space of $A$, that is, of $\{A_1, A_2, \ldots, A_n \}$; thus (11) binds.
We now apply these considerations to $xy^\ast$ and immediately deduce via (10) that
$\text{rank}(xy^\ast) = \dim \text{range}(xy^\ast) = \dim \text{span} \{x\} = 1, \tag{15}$
as per request. End of Note.