Diode - draw output waveform
Let me redraw your schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

It's a lot easier to recognize this way. The diodes are arranged to cause clipping at the common node shared by \$D_1\$, \$D_2\$, \$R_1\$, and \$C_1\$ -- namely \$V_a\$.

The following "graph" shows you the input voltage in blue and the \$V_a\$ node voltage in red:







simulate this circuit

Note that it follows the input voltage until one, or the other, diode conducts. At which time, it stays at the appropriate battery voltage on the other side of that diode. This means that the positive peak is at \$+6\:\textrm{V}\$ and the negative peak is at \$-4\:\textrm{V}\$. This is the clipping effect at node \$V_a\$.



Now, there is missing information from your problem: the initial voltage across the capacitor. But when a problem doesn't state it, then we can fairly argue that it is uncharged and has zero volts across it.

As a consequence of this "assumption," for the first cycle only the voltage on the other side (right side) of your capacitor should "follow" the curve in red shown above. That is, if there wasn't a diode present at the output. But there is. So this is yet another problem to consider.

If you examine the red curve above and assume that there is no voltage across the capacitor, then the output will be able to follow the positive portion of the red curve since the output diode is reverse-biased.

That portion does not look like the curve you provided.

On the negative portion of the red curve, however, the diode is forward biased and therefore conducts. This means that the output is prevented from going more negative (in the ideal-diode case) than zero volts.

That portion does look more like the curve you provided. (Not entirely, though, as we'll soon see.)



However, there's more.

While the red curve reaches its most negative peak value of \$-4\:\textrm{V}\$, the capacitor's voltage is fully charged up to that difference, since the diode is conducting and holding the output to zero volts.

Therefore, immediately at the point where the red curve's voltage begins rising back up towards zero volts, the capacitor will have its left side negatively charged (which was just at \$-4\:\textrm{V}\$, relative to its right side (which was just at \$0\:\textrm{V}\$.) So as the red curve moves back upwards on that final leg, the capacitor voltage will add to it. Because of the capacitor polarity, this will add \$+4\:\textrm{V}\$ to the final leg of the red curve.

As the last final leg of the red curve rises upward, the output will be \$+4\:\textrm{V}\$ higher (meaning it will rise above zero volts) and the output diode will cease to conduct (it will be off.) Therefore the output will rise upward with the last leg of the red curve, but \$+4\:\textrm{V}\$ higher.



This should be more than sufficient information for you to compose a final output diagram for the first cycle.



However, there is still more.

Now. If you are supposed to provide the steady-state case, after the capacitor gets a charge from the first cycle, the answer will be a little different. From your comments that I now see, you do need to show the next cycle (or steady-state case.)

So you need to follow up by realizing that after the first cycle the capacitor will be charged up. Apply this knowledge to the red curve I provided but now with the added voltage to it.

Note that node \$V_a\$ will still look the same in the next cycle (verify this in your own mind.) However, now you will have a charged capacitor to deal with in working out what \$V_{out}\$ looks like.

See where that takes you.