Proving a set of linear functionals is a basis for a dual space I've seen some similar problems on the stackexchange and I want to be sure I am at least approaching this in a way that is sensible. 
The problem as stated: 
Let $V= \Bbb R^3$ and define $f_1, f_2, f_3 \in V^*$ as follows: 
$f_1(x,y,z)= x-2y ,\; f_2(x,y,z)= x+y+z,\;f_3(x,y,z)= y-3z$
Prove that $f_1, f_2, f_3$ is a basis for $V^*$ and and then find a basis for V for which it is the dual basis. 
Here's my problem: the question feels a bit circular. But this is what I attempted: 
To show that the linear functionals $f$ are a basis, we want that $f_i(x_j)=\delta_{ij}$, or that $f_i(x_j)=1$ if $i=j$ and that it is zero otherwise. 
That means that  we want to set this up so that 
$$1= f_1(x,y,z)= x-2y$$ 
$$0= f_2(x,y,z)= x+y +z$$ 
$$0= f_3(x,y,z)=   y-3z$$
That gives us three equations and three unknowns. Solving them we get $2x-2z=1$ for $x-z=\frac{1}{2}$ and $z=x-\frac{1}{2}$ and subbing into the equation for $f_3$ I get $0=y-3x-\frac{3}{2}$ which gets us $1=x-6x+3$ or $x=\frac{2}{5}$. That gives us $y=\frac{-3}{10}$ and $z=\frac{-1}{10}$. 
OK, this is where I am stuck on the next step. I just got what should be a vertical matrix I think, with the values $(\frac{2}{5}, \frac{-3}{10}, \frac{-1}{10})$ but I am not sure where to go from here. I am not entirely sure I set this up correctly. 
thanks
EDIT: I do know that I have to show that $f_1, f_2, f_3 $ are linearly independent. That I think I can manage, but I am unsure how to fit it into the rest of the problem or if I am even approaching this right. 

What about a direct approach? Suppose $\,a,b,c\in\Bbb R\,$ are such that
$$af_1+bf_2+cf_3=0\in V^*\implies\;\forall\,v:=(x,y,z)\in\Bbb R^3\;,\;\;af_1v+bf_2v+cf_3v=0\iff$$
$$a(x-2y)+b(x+y+z)+c(y-3z)=0\iff$$
$$\iff (a+b)x-(2a-b-c)y+(b-3c)z=0$$
As the above is true for all $\;x,y,z\in\Bbb R\,$ , we must have
$$\begin{align*}\text{I}&\;\;\;\;a+b=0\\\text{II}&\;\;\;\;2a-b-c=0\\\text{II}&\;\;\;\;b-3c=0\end{align*}\;\;\implies a=-b\;,\;\;c=3a=\frac13b\implies a=b=c=0$$
and we're done.
