Oh Jesus! If you take two bases $E$ and $F$, the matrix of change from $E$ to $F$ is given by: $$M_{EF} = [{\rm id}]_{FE}.$$Then $E$ and $F$ have the same orientation if $\det M_{EF} > 0$. Note that this determinant won't ever be zero (why?). What you did is valid, but your notation was sloppy: ${\bf e}_1 = (1,0,0)$ is being used to mean $[{\bf e}_1]_E = (1,0,0)$, etc - the vector ${\bf e}_1$ is not necessarily the vector $(1,0,0) = {\bf i}$.
A sure-fire way to do this is to compute $[{\rm id}]_{EF}$. $$\begin{cases} {\rm id}({\bf f}_1) = {\bf f}_1 = 1\cdot {\bf e}_1 + 0 \cdot {\bf e}_2 + 0\cdot {\bf e}_3 \\ {\rm id}({\bf f}_2) = {\bf f}_2 = 0\cdot {\bf e}_1 + 1 \cdot {\bf e}_2 + 1\cdot {\bf e}_3 \\ {\rm id}({\bf f}_3) = {\bf f}_3 = 1\cdot {\bf e}_1 + 1 \cdot {\bf e}_2 + 0\cdot {\bf e}_3 \end{cases}$$
Building the matrix: $$M_{EF} = [{\rm id}]_{FE} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix} \implies \det M_{EF} = -1 <0,$$ so they have opposite orientations. Note, though, that "positive" and "negative" orientations only make sense w.r.t. a fixed basis in space - usually the canonical basis. If you define $E \sim F$ if and only if $\det M_{EF} > 0$, you get an equivalence relation. Since the determinant is always positive or negative, we only get two equivalence classes. Pick the class of the canonical basis and call it "positive".
Let $\mathbf{x} = (x_{1}, x_{2}, x_{3})^T$ be the coordinates of a point in the $e$-basis and let $\mathbf{y} = (y_{1}, y_{2}, y_{3})^T$ be the coordinates of the same point in the $f$-basis.
It is the same point, so we require the following condition.
$$
x_{1} \mathbf{e}_{1} +
x_{2} \mathbf{e}_{2} +
x_{3} \mathbf{e}_{3}
=
y_{1} \mathbf{f}_{1} +
y_{2} \mathbf{f}_{2} +
y_{3} \mathbf{f}_{3}
$$
The question gives the way of writing the $f$-basis vectors in terms of the $e$-basis vectors:
$$
\begin{aligned}
\mathbf{f}_1 &= \mathbf{e}_1 + \mathbf{e}_2 \\
\mathbf{f}_2 &= \mathbf{e}_2 \\
\mathbf{f}_3 &= \mathbf{e}_1 - \mathbf{e}_3
\end{aligned}
$$
We can substutite these formulas into the equation for the coordinates above
$$
x_{1} \mathbf{e}_{1} +
x_{2} \mathbf{e}_{2} +
x_{3} \mathbf{e}_{3}
=
y_{1} (\mathbf{e}_{1} + \mathbf{e}_{2}) +
y_{2} \mathbf{e}_{2} +
y_{3} (\mathbf{e}_{1} - \mathbf{e}_{3})
$$
$$
x_{1} \mathbf{e}_{1} +
x_{2} \mathbf{e}_{2} +
x_{3} \mathbf{e}_{3}
=
y_{1} \mathbf{e}_{1} + y_{1} \mathbf{e}_{2} +
y_{2} \mathbf{e}_{2} +
y_{3} \mathbf{e}_{1} - y_{3} \mathbf{e}_{3}
$$
$$
x_{1} \mathbf{e}_{1} +
x_{2} \mathbf{e}_{2} +
x_{3} \mathbf{e}_{3}
=
(y_{1} + y_{3}) \mathbf{e}_{1}
+
(y_{1} + y_{2} ) \mathbf{e}_{2}
- y_{3} \mathbf{e}_{3}
$$
Now $\mathbf{e}_1$, $\mathbf{e}_2$ and $\mathbf{e}_3$ are three linearly independent vectors so that the components of each vector can be equated on both sides of the above. I.e. we can write:
$$
\begin{aligned}
x_1 &= y_1 + y_3 \\
x_2 & = y_1 + y_2 \\
x_3 &= -y_3
\end{aligned}
$$
The following is exactly the same as the above with slightly different spacing
$$
\begin{aligned}
x_1 &= y_1 & &+y_3 \\
x_2 & = y_1 &+ y_2 & \\
x_3 &= & &-y_3
\end{aligned}
$$
Writing the equations that relate the coordinates in this way, we can see how the set can be written as a single matrix equation:
$$
\begin{pmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 0 & -1
\end{pmatrix}
\begin{pmatrix}
y_{1} \\ y_{2} \\ y_{3}
\end{pmatrix}
\Rightarrow
\mathbf{x}
=
\begin{pmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 0 & -1
\end{pmatrix}
\mathbf{y}$$
This shows how we can use a matrix to convert coordinates in the $f$-basis to coordinates in the $e$-basis, i.e. $\mathbf{x} = T \mathbf{y}$, i.e. it represents the matrix $T$ in the formula
$$
A' = T^{-1} A T
$$
where $A$ is the transformation that is applied to coordinates in the $e$-basis. The above formula is applied
Having found $T$, we can find its inverse (by hand or with some software):
$$
T^{-1}
=
\begin{pmatrix}
1& -1& 0 \\
0& 1& 0 \\
1& -1& -1
\end{pmatrix}
$$
and finally, we can calculate $A'$
$$
A'
=
\begin{pmatrix}
-3& -2& -2 \\
3& 3& -1 \\
-7& -1& -6
\end{pmatrix}
$$
which is the matrix of the transformation that is applied to coordinates in the $f$-basis.
This next part goes into how the formula relating the two transformation matrices in the different bases is derived.
If we write $\mathbf{u}$
for the result of applying $A$ to the $e$-basis vector $\mathbf{x}$ and if we write
$\mathbf{v}$ for the result of applying $A'$ to the $f$-basis vector $\mathbf{y}$.
$$
\begin{aligned}
\mathbf{u} &= A \mathbf{x} \\
\mathbf{v} &= A' \mathbf{y} \\
\end{aligned}
$$
The vectors
$\mathbf{x}$ and $\mathbf{y}$
correspond to the same point in the two different bases and so do the pair of vectors
$\mathbf{u}$ and $\mathbf{v}$. In other words they can be written:
$$
\begin{aligned}
\mathbf{x} &= T \mathbf{y} \\
\mathbf{u} &= T \mathbf{v} \\
\end{aligned}
$$
This means we can write the following
$$
\begin{aligned}
\mathbf{u} &= T \mathbf{v} \\
A \mathbf{x} &= T \mathbf{v} \\
A \mathbf{x} &= T A' \mathbf{y} \\
A T\mathbf{y} &= T A' \mathbf{y} \\
T^{-1} A T\mathbf{y} &= A' \mathbf{y} \\
\end{aligned}
$$
As this works for all $\mathbf{y}$ we can conclude that $T^{-1} A T = A' $.
Best Answer
Your work looks good to me. Here's another way to do it (just for fun).
First recognize that $(1,0,0) = \frac 12e_1 + \frac 12 e_3$
Thus $$f_1(x) = f_1\left(\frac 12e_1 + \frac 12 e_3\right) = \frac 12f_1(e_1) + \frac 12f_1(e_3) = \frac 12 (1) + \frac 12 (0) = \frac 12 \\ f_2(x) = 0f_2(e_2) = 0 \\ f_3(x) = \frac 12f_3(e_3) = \frac 12$$
(For simplicity I just ignored the irrelevant terms in the last two lines.)