Am I correctly finding the standard matrix? 
Question:
Let $F: \Bbb R^3 \to \Bbb R^3$ be the linear transformation satisfying
  $F(1,0,1)=(-3,-3,1)$,$F(0,1,0)=(0,1,1)$, and $F(0,1,1)=(2,-2,1)$. Find
  the standard matrix $A$ of $F$.

My Approach:
I used a method that I haven't been taught, so I'm not sure if I am correct and allowed to do this. Can you guys see if this is acceptable.
Since:
$$F\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}3\\-3\\1\end{pmatrix}$$
$$F\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\1\end{pmatrix}$$
$$F\begin{pmatrix}0\\1\\1\end{pmatrix}=\begin{pmatrix}2\\-2\\1\end{pmatrix}$$
We want to find:
$$\text{standard matrix:}F\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=?$$ 
So what I did was:

$$\therefore \text{standard matrix:}F\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}3&0&2\\-1&0&-1\\-2&1&-1\end{pmatrix}$$
Is this method correct?

Your matrix does not map $(0,1,0)^t$ to $(0,1,1)^t$, so I believe it is wrong.
Instead you could use
$$
A X = Y \iff \\
A = Y X^{-1}
$$
where $X$ has the given argument vectors and $Y$ the corresponding image vectors, assuming the argument vectors are linear independent.
Applying the above I get
$$
X =
\left(
\begin{array}{rrr}
1 &0 & 0 \\
0 &1 & 1 \\
1 &0 & 1
\end{array}
\right)
\quad\quad
Y=
\left(
\begin{array}{rrr}
 3 & 0 &  2 \\
-3 & 1 & -2 \\
 1 & 1 &  1
\end{array}
\right)
$$
which leads to
$$
A=
\left(
\begin{array}{rrr}
1&0&2\\
0&1&-3\\
1&1&0
\end{array}
\right)
$$
The matrix you calculated is $B = X^{-1} Y$. It solves $X B = Y$.
