Finding a Topological Conjugacy Using Hartman-Grobman Here is the problem statement:
Find a homeomorphism $h$ in a neighborhood of $0$ that establishes a topological conjugacy between the flow of the differential system and the flow of the linearized system, i.e.,
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad h(\psi(t, x)) = e^{tA}h(x)$
where $A = Df(0)$ and $\boldsymbol{\psi}(t, x_0)$ is the solution of $\mathbf{\dot{x}} = f(\mathbf{x})$ and $\mathbf{x}(t_0) = x_0$, the nonlinear system given by
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad
\dot{x} = −x \\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\dot{y} = −y + xz \\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\dot{z}=z .
$
Without loss of generality, I let $t_0=0$ and found that the solutions for both the linear and non-linear systems, respectively, should be
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\mathbf{x}=\begin{pmatrix} x_0e^{-t} \\ y_0e^{-t} \\ z_0e^t \end{pmatrix}$
and
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\boldsymbol{\psi} = \begin{pmatrix} x_0e^{-t} \\ x_0z_0+(y_0-x_0z_0)e^{-t} \\ z_0e^t \end{pmatrix} .$
We can calculate the matrix exponential since $A=\begin{pmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{pmatrix}$:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad e^{tA} = \begin{pmatrix} e^{-t} & 0 & 0 \\ 0 & e^{-t} & 0 \\ 0 & 0 & e^t \end{pmatrix} .$
I think that my misunderstanding arises from my confusion in what is required of the homeomorphism $h$. Must $h$, in fact, be a matrix transformation which is time-independent? Also, in the right-hand side of the above equality I wish to show with $h$, should $x=\mathbf{x}$, the solution of the linear system? What I understand is that I am showing that flowing in the non-linear space and mapping by $h$ is the same as mapping by $h$ and then flowing in the linear space.

It might be easier to rename the variables so that the nonlinear system still lives in $xyz$-space, but the linear system lives in $uvw$-space instead.
So you have the solution of the nonlinear system,
$$
(x(t),y(t),z(t)) = (x_0 \, e^{-t}, x_0 z_0 + (y_0 - x_0 z_0) \, e^{-t}, z_0 \, e^t)
,
$$
and the solution of the linear system,
$$
(u(t),v(t),w(t)) = (u_0 \, e^{-t}, v_0 \, e^{-t}, w_0 \, e^t)
. 
$$
What you want to find is a function
$$
(u,v,w)=h(x,y,z)
$$
(independent of $t$) such that
$$
h(x(t),y(t),z(t))
$$
is the same thing as
$$
(u(t),v(t),w(t))
\qquad
\text{with}
\quad
(u_0,v_0,w_0) = h(x_0,y_0,z_0)
.
$$
Now, the first thing one notices when comparing the nonlinear and linear solutions is that $x(t)$ and $z(t)$ are doing the same thing as $u(t)$ and $w(t)$, which suggests that we can just map $x$ and $z$ to $u$ and $w$:
$$
(u,v,w) = h(x,y,z) = (x, [?], z)
.
$$
To figure out what's going to be in the middle slot we need to be a bit more clever. But we may notice that in the nonlinear solution, the quantity $y(t) - x(t) z(t)$ is just decreasing exponentially from $y_0-x_0 y_0$ to $0$, as $t$ goes from $0$ to $\infty$. And this is similar to what $v(t)$ is doing in the linear solution: it decreases exponentially from $v(0)$ to $0$. So we could try with
$$
(u,v,w) = h(x,y,z) = (x, y-xz, z)
,
$$
perhaps?
I'll leave it to you to check that this really works (and that it's actually a global diffeomorphism, not just a local homeomorphism).
(One may also view it as follows: since $x(t) z(t) = x_0 z_0$ is constant, we have $(xz)'=0$, so the nonlinear system can be written as
$$
x'=-x
,\quad
(y-xz)'=-(y-xz)
,\quad
z'=z
,
$$
and thus the change of variables $u=x$, $v=y-xz$, $w=z$ actually performs the linearization.)
