TL;DR: I wasn't going to bother doing a full derivation and settle with a handwavy argument that we know the load current is zero because the two sub-circuits are identical and so the voltage across the load resistor will be zero, but I decided that it would probably not help very much, so full derivation it is.
Lets ignore the ground node that you have placed - there is only one of it, so it is doing nothing. Also, lets label the node between \$R_1\$ and \$R_2\$ as \$V_u\$, and the node between \$R_3\$ and \$R_4\$ as \$V_w\$.
It is then worth a reminder of what a potential difference is. It is the difference in voltage between any two nodes. The 9V battery has a potential difference at its positive terminal of 9V referenced to its negative terminal.
Lets first consider the sub-circuit individually - you can't always do this, in fact frequently you can't, but because the two are identical, I know that I can. You have 6V dropped across effectively a 1mOhm resistor. As a result you are going to get large currents - 6kA worth.
Lets try to work out the voltage at the node \$V_u\$ that I defined earlier with respect to node \$V_x\$, or to put it another way, the potential difference \$V_{ux}\$.
Well, we know from Kirchoffs Current Law that the sum of all currents on all braches at a node equals zero. So what are the currents into the node U? Well, there is the load current, the current through \$R_1\$ and the current through \$R_2\$. Lets put this in a calculation:
$$I_{R_1} + I_{R_2} + I_{load} = 0$$
So what are these currents. Well, lets work them out. First \$I_{R_1}\$
$$I_{R_1} = \frac{V_{R_1}}{R_1} = \frac{V_u - (V_x-9)}{R_1} = \frac{V_u - V_x + 9}{500\mu\Omega} $$
Then \$I_{R_2}\$
$$I_{R_2} = \frac{V_{R_2}}{R_2} = \frac{V_u - (V_x-3)}{R_2} = \frac{V_u - V_x + 3}{500\mu\Omega} $$
Bringing that all together:
$$I_{load(u)} = -\frac{V_u - V_x + 3}{500\mu\Omega} - \frac{V_u - V_x + 9}{500\mu\Omega} = \frac{-(V_u - V_x + 3) - (V_u - V_x + 9)}{500\mu\Omega} = \frac{-2V_u + 2V_x - 12)}{500\mu\Omega} = \frac{V_x - V_u -6}{1m\Omega}$$
Do the same with the other subsystem - you will get the same equation but with \$V_y\$ and \$V_w\$ instead - i.e:
$$I_{load(v)} = \frac{V_y - V_w -6}{1m\Omega}$$
Now we have analysed each subsystem, lets calculate what the load current is. We know that the current flowing through load resistor will be equal in magnitude on each side - because currents through a series branch are always equal.
$$\begin{align}
I_{load(u)} &= I_{load(v)}\\\\
\frac{V_x - V_u -6}{1m\Omega} &= \frac{V_y - V_w -6}{1m\Omega}\\\\
V_x - V_u &= V_y - V_w \tag{1}\\
\end{align}$$
If we look at the connection between \$V_x\$ and \$V_y\$ we can see that they are directly connected. That is to say \$V_x = V_y\$. So, substituting that in to (1), we get:
$$\begin{align}
V_y - V_u &= V_y - V_w\\\\
V_u &= V_w \tag{2}\\
\end{align}$$
We know then work out the current through the load is:
$$I_{load} = \frac{V_{load}}{R_{load}} = \frac{V_{uw}}{2k\Omega}= \frac{V_{u}-V_{w}}{2k\Omega} \tag{3}$$
So substituting (2) into (3) we get:
$$I_{load}= \frac{0}{2k\Omega}=0A$$
We can see then that there is no current through the load.
You can also then work out the current flowing in each sub-circuit. Remember we now know that the load current is zero which means each sub-circuit is effectively independent. You should be able to work out that the voltage over \$R_1\$ and \$R_2\$ is \$9-3 = 6V\$. So we can say that the current is:
$$I = \frac{6}{R_1+R_2} = \frac{6}{1m\Omega}=6kA $$