Calculate current over ideal diode
I'd start with the assumption that both diodes are forward biased and see what happens.

D1 connected to ground means the anode is 0.7V. D2 will have a forward drop of 0.7 meaning that it's actually creating a virtual ground at V(out).

Then it's just voltages over resistors. (5-0.7)/10k = current through 10k. (0-(-5))/5k = current through 5k.

Then use KCL to determine what I is. Current through 10k flows into node, current through 5K flows out of node, and current I flows out of node. So you should have an equation like this:
$$I_{10k}-I_{5k}-I=0$$

When you do this, you'll see that I is actually negative which can't be because the diode would then be reversed biased and would block all current. That means D1 becomes an open in this circuit.

Now you just have one series line of voltage across resistors and a diode. Ohms law states this:
$$\frac{5-(-5)-0.7}{10k+5k}=I_{series}$$

With Iseries and 5k, you can find Vout. You know that I is 0 because D1 is reverse biased.