Answer
$i_{1}=i_{2}=-0.5A$
Work Step by Step
In this question, we are required to find the two currents $i_{1}$ and $i_{2}$ by using node voltage analysis, i.e. considering the differences in potential across the nodes in the circuit.
By KCL (Kirchoff's Current Law), $\sum{i_{n}}=0$. Since $1A$ of current approaches the node to the left of $R_{2}$ and $i_{1}A$ of current leaves at the node, $(1-i_{1})A$ of current continues rightwards and passes through $R_{2}$.
Since $R_{2}=1\Omega$,
$({\rightarrow}^{+})$ P.D. (Potential Difference) across $R_{2}=(1-i_{1})V$
Now consider the node above $R_{3}$.
$({\downarrow}^{+})$ P.D. across $R_{3}=6i_{2}V$
Now consider the node above $R_{1}$.
P.D. across $R_{1}=3i_{1}V$
Since $R_{2}$ and the joint directly below it are in parallel, both joints have the same P.D. P.D. at the bottom can be expressed as $3i_{1}-6i_{2}$.
$$ \therefore {3i_{1}}-6i_{2}=1-i_{1}$$
$$4i_{1}-6i_{2}=1$$
We also note that the $1A$ and $2A$ currents flow through empty joints between the nodes across $R_{1}$ and $R_{2}$. This indicates that an equivalent imperfect voltage source of $3i_{1}V$ with an internal resistance of $3\Omega$ generates $1A$ of current, while an imperfect voltage source of $6i_{2}V$ with an internal resistance of $6\Omega$ generates $2A$ of current.
We thus have the two equations
$$\frac{3i_{1}}{k}=1$$
$$\frac{6i_{2}}{k}=2$$
where $k\in\mathbb{R}$.
(Note: The intention here is to express $i_{1}$ and $i_{2}$ in numerical ratios and compare them, so that we can express one variable in terms of the other. To do this, we peg a "placeholder" variable $k$ to the resistance of the wire. Since we are only concerned with taking ratios in $3i_{1}$ and $6i_{2}$, $k$ will be eliminated in the ratios and thus, we do not have to worry about whether $k$ will affect subsequent calculations depending on whether we assume the wires are ideal or not.)
We thus find that $i_{1}=\frac{1}{3k}$ and $i_{2}=\frac{2}{6k}=\frac{1}{3k}$.
$$\therefore\frac{i_{1}}{i_{2}}=1\Rightarrow{i_{1}}=i_{2}$$
Substituting $i_{1}=i_{2}$ into the equation we derived earlier,
$$4i_{2}-6i_{2}=-2i_{2}=1\Rightarrow{i_{2}}=-0.5A$$
$$\therefore{i_{1}}=i_{2}=-0.5A$$
