Writing
$z = x + iy, \tag 1$
where $x$, $y$ and $z$ are functions of the independent variable $t$, we have
$\dot z = \dot x + i \dot y, \tag 2$
and
$z^\ast = x - iy; \tag 3$
then the given equation
$\dot z = z^\ast \tag 4$
becomes
$\dot x + i \dot y = x - iy; \tag 5$
equating the real and imaginary parts we find
$\dot x = x \tag 6$
and
$\dot y = -y; \tag 7$
to proceed further, we assign $z$ the initial condition
$z(t_0) = x(t_0) + iy(t_0), \tag 8$
which in fact initializes $x$ and $y$; now (6) and (7) have the well-known unique solutions
$x(t)= x(t_0)e^{t - t_0}, \tag{9}$
$y(t) = y(t_0)e^{t_0 - t}, \tag{10}$
whence
$z(t) = x(t_0)e^{t - t_0} + iy(t_0)e^{t_0 - t}. \tag{11}$
In order to sketch the phase portrait of the system (4), we first take note of the fact that (9) and (10) together yield
$x(t)y(t) = x(t_0)y(t_0), \tag{12}$
which whenever
$x(t_0) \ne 0 \ne y(t_0) \tag{13}$
shows that the solution curves lie in he hyperbolas
$xy = x(t_0)y(t_0) \tag{14}$
in the $xy$-plane $\Bbb R^2$; if precisely one of $x(t_0)$, $y(t_0)$ does not vanish, then the corresponding phase curve is either the positive or negative $x$ or $y$ half-axis. For example, if
$x(t_0) > 0, \; y(t_0) = 0, \tag{15}$
the solution curve traverses the entire positive $x$-axis as $-\infty \to t \to \infty$. Similar results apply when
$x(t_0) = 0, \; y(t_0) \ne 0 \tag{16}$
etc; the reader may easily complete the necessary details; finally when
$x(t_0) = 0 = y(t_0) \tag{17}$
the phase curve is the single point $(0, 0)$. With these facts in mind, the phase portrait of (4) may be sketched with little additional effort.
Best Answer
With $z'=e^{it}\bar z$ you also get by conjugation $\bar z'=e^{-it}z$ and thus for the second derivative $$z'' = ie^{it}\bar z+e^{it}\bar z'=iz'+z.$$ This second order linear ODE has as characteristic polynomial $$ \lambda^2-iλ-1 =\left(λ-\frac i2\right)^2-\frac34 =\left(λ-\frac{i-\sqrt3}2\right)\left(λ-\frac{i+\sqrt3}2\right) $$ which allows you to construct the solution. $$ z=e^{it/2}\left(c_1e^{\sqrt3 t/2}+c_2e^{-\sqrt3 t/2}\right)\\ $$
You have to find a relation between the integration constants that restricts the general solution to the solutions of the original equation. \begin{align} z'&=e^{it/2}\left(\frac12(i+\sqrt3)c_1e^{\sqrt3 t/2}+\frac12(i-\sqrt3)c_2e^{-\sqrt3 t/2}\right) \\ z'-e^{it}\bar z&=e^{it/2}\left(\frac12\Bigl[(i+\sqrt3)c_1-2\bar c_1\Bigr]e^{\sqrt3 t/2}+\frac12\Bigl[(i-\sqrt3)c_2-2\bar c_2\Bigr]e^{-\sqrt3 t/2}\right) \end{align} This implies $$\Bigl[(\sqrt3+1)+(\sqrt3-1)i\Bigr]c_1=\Bigl[(\sqrt3+1)-(\sqrt3-1)i\Bigr]\bar c_1$$ where the right side is the conjugate of the left and thus both sides are real, so that $$c_1=A\,\Bigl[(\sqrt3+1)-(\sqrt3-1)i\Bigr]$$ and similarly $$c_2=B\,\Bigl[(\sqrt3-1)-(\sqrt3+1)i\Bigr]$$ with real constants $A,B\in \Bbb R$.
As the components have absolute values $e^{\pm\sqrt3/2\,t}$, you get that all non-trivial solutions are unbounded over $\Bbb R$.