I have a circuit here. For negative output voltages from \$V_1\$, the diode will be off, making \$V_0\$ simple to calculate since the circuit just becomes resistors. When \$V_1 = 0\$ the diode is off since no current flows.
Using the simplified diode model (fixed Vf of 0.7 volts, say), you can see there are two possible situations- diode conducting and diode not conducting. When the diode is not conducting you replace it with an open circuit and analyze easily.
When the diode is conducting, replace it with a voltage source of Vf, and the analysis is not much more difficult.
It should be easy to see that the diode just starts to conduct when the forward voltage is exactly equal to Vf without it conducting - so the first analysis will give you that point. Since there is no current, the voltage across R2 is 0, and the equation is just Vf= Vi*6/7.
So you have:
case 1: Vo = f(Vi) valid for Vi <= 7/6 * Vf
case 2: Vo = f'(Vi) valid for Vi >= 7/6*Vf
If your input voltage is always > 1V you may only have to deal with one situation.
You have a few options here for the diode when conducting, in increasing order of precision (not necessarily effort):
Constant-voltage model. Assume that the voltage across it is a constant voltage \$V_f\$. That is, when the diode is forward conducting, assume it is a voltage source with anode = positive terminal. This model is only valid for a certain range of currents \$I_D\$ at which \$V_f\$ was measured (often specified in datasheet). We often estimate that a silicon diode has a forward voltage of about 0.5 to 0.7V (depending on currents; up to 1.2V for high currents), and a schottkey diode around 0.3V.
Estimate the current through the diode, and use the V-to-I graph in the diode datasheet to find the voltage across the diode for that specific current. Assume the constant voltage model using this value of voltage. This is useful when the \$V_f\$ specified in the datasheet is for a current wildly different from what you will actually see, but the currents you expect in the circuit remain in a small range. (Or when accuracy doesn't matter too much, but you want to keep it simple.)
Do hand analysis with the symbol \$V_D(I_D)\$ (diode voltage as a function of diode current). Use numerical analysis methods, along with the V-to-I graph in the diode datasheet, to solve the circuit equations. You might be able to figure out a way of implementing the secant method using the graph. Or you can do something like guess at \$V_D\$ (for example, 0.6V), get \$I_D\$ from the graph, re-calculate \$V_D\$ from the hand-analysed equations, etc. You should find after a few iterations that this converges.
Same as 3, but fit the graph to the Shockley diode equation. Then you also have the option of using software to solve the circuit for you, and also open up the Newton-Raphson method, rather than having to do work by hand. This also opens you up to hand analysis—usually this ends up being a set of transcendental equations that aren't possible to solve by hand without introducing a new function, so you end up doing the iterative method from #3 with two equations instead of one equation and one graph.
Use a circuit simulator and load up the SPICE model for the diode you're using.
Side note: if you're interested in more accurately modelling the diode in reverse bias, consider it leaking a constant current (that is, it's a constant current source). The reverse leakage current can also be found in the datasheet.