First, you should verify that the $f_i$ are elements of $V^*$; that is, that they are functions on $\Bbb R^3$ that satisfy:
$\ \ \ $1) $f({\bf x}+{\bf y})=f({\bf x})+f({\bf y})$ for all ${\bf x},{\bf y}\in\Bbb R^3$
and
$\ \ \ $2) $f(c{\bf x})=cf({\bf x})$ for all $c\in\Bbb R$ and all ${\bf x}\in \Bbb R^3$.
Of course, if you know this has been covered in your class already, you're probably safe just writing something like "as was demonstrated in lecture blah, these are elements of the dual of $V=\Bbb R^3$.
Another fact I assume you have use of is that the dimension of $V^*$ is three (as the dimension of the vector space $\Bbb R^3$ is three).
So, with three linear functionals on $\Bbb R^3$, towards showing that they are a basis of $V^*$, it suffices to show that they are independent. There are many ways towards achieving this end. One way in particular is to show that the matrix $A$ formed by taking as its rows the coefficients of the $f_i$, $ A=\left[\matrix{1&-2&0\cr 1&1&1\cr 0&1&-3}\right], $ has full rank (that this is so isn't hard to see: $c_1f_1+c_2f_2+c_3f_3={\bf 0}\iff A\bigl[{{\scriptstyle c_1\atop\scriptstyle c_2}\atop c_3}\bigr]=\bf 0$; and the former equation has only the trivial solution if and only if $A$ has full rank).
So, let's row reduce $A$: $ A= \left[\matrix{1&-2&0\cr 1&1&1\cr 0&1&-3}\right] \buildrel{r_2-r_1\rightarrow r_2}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 3& 1\cr 0&1&-3}\right] \buildrel{r_2-3r_3\rightarrow r_3}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 3& 1\cr 0&0&10}\right]. $
At this point we can see that $A$ indeed has full rank. Thus the set $\{f_1,f_2,f_3\}$ is independent and consequently is a basis of $V^*$.
Towards finding the dual basis let's recall what this is: the dual basis of $\{f_1,f_2,f_3\}$ by definition is the basis $\{ {\bf x}_1,{\bf x}_2,{\bf x}_3\}$ of $V$ for which $ f_i({\bf x}_j)=\cases{1,& if $i=j$\cr 0, & if $i\ne j$}. $
So, in particular, the first basis element of dual basis, ${\bf x}_i$, would satisfy $ f_1({\bf x}_1)=1,\ f_2({\bf x}_1)=0, \text{and}, f_3({\bf x}_1)=0. $ In other words, we have the system: $ \eqalign{ x-2y&=1\cr x+y+z&=0\cr y-3z&=0 } $ whose matrix form is $ A {\bf x}_1=\left[\matrix{ 1\cr0\cr 0 }\right] $ and whose, necessarily unique, solution gives the coordinates of ${\bf x}_1$.
We would have two similar systems to solve in order to find ${\bf x}_2$ and ${\bf x}_3$.
That seems like a lot of work; but wait... suppose we wrote the dual basis as a matrix whose columns were the ${\bf x}_i$. Then we'd have: $ A[ {\bf x}_1\ {\bf x}_2\ {\bf x}_3] =\left[\matrix{1&-2&0\cr 1&1&1\cr 0&1&-3}\right][ {\bf x}_1\ {\bf x}_2\ {\bf x}_3]=\left[\matrix{1&0&0\cr 0&1&0\cr 0&0&1}\right] $
So $[ {\bf x}_1\ {\bf x}_2\ {\bf x}_3]$ is the inverse of $A$. Rather than solving three systems of equations, we could instead find the inverse of $A$ and then the columns will be our dual basis.
This is what we'll do. Towards that end, we may (and do) adjoin the identity matrix to $A$ and perform a full forward/back row reduction:
$\eqalign{ [A \,|\,I\,]= \left[\matrix{1&-2&0\cr 1&1&1\cr 0&1&-3} \ \ \Biggl|\ \ \matrix{1& 0&0\cr 0&1&0\cr 0&0&1}\right] &\buildrel{r_2-r_1\rightarrow r_2}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 3& 1\cr 0&1&-3} \ \ \Biggl|\ \ \matrix{1& 0&0\cr -1&1&0\cr 0&0&1}\right]\cr &\buildrel{r_2-3r_3\rightarrow r_3}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 3& 1\cr 0&0&10} \ \ \Biggl|\ \ \matrix{1& 0&0\cr -1& 1&0\cr -1&1&-3}\right]\cr &\buildrel{10r_2-r_3\rightarrow r_2}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 30& 0\cr 0&0&10} \ \ \Biggl|\ \ \matrix{1& 0&0\cr -9&9&3\cr -1&1&-3}\right]\cr &\buildrel{15r_1+r_2\rightarrow r_1}\over{\longrightarrow} \left[\matrix{15&0&0\cr 0& 30& 0\cr 0&0&10} \ \ \Biggl|\ \ \matrix{ 6& 9&3\cr -9&9&3\cr -1&1&-3}\right]\cr &\buildrel{ }\over{\longrightarrow} \left[\matrix{1 &0&0\cr 0& 1& 0\cr 0&0&1 } \ \ \Biggl|\ \ \matrix{ 6/15& 9/15&3/15\cr -9/30&9/30&3/30\cr -1/10&1/10&-3/10}\right].\cr } $
So $ A^{-1}=\left[\matrix{2/5& 3/5&1/5\cr -3/10&3/10&1/10\cr -1/10&1/10&-3/10}\right], $ and the dual basis has as its elements the columns of $A^{-1}$: $ {\bf x}_1=\left[\matrix{2/5\cr-3/10\cr-1/10 }\right],\ {\bf x}_2=\left[\matrix{3/5\cr3/10\cr1/10 }\right],\ {\bf x}_3=\left[\matrix{1/5\cr1/10\cr-3/10 }\right]. $
The basis $\{ {\bf x}_1,{\bf x}_2,{\bf x}_3 \}$ is the dual basis of $\{f_1,f_2,f_3\}$.
Note that the relative ordering is important. For instance, we must have $f_3({\bf x}_1)=f_3({\bf x}_2)=0$ and $f_3({\bf x}_3)=1$. As a spot check, let's verify this: $\eqalign{ f_3({\bf x}_1)&= (-3/10)-3(-1/10)=0\cr f_3({\bf x}_2)&= (3/10)-3(1/10)=0\cr f_3({\bf x}_3)&= (1/10)-3(-3/10)=1.\cr } $