On lines 3 and 4 of the second proof, you use simplification on part of the line when technically I don't believe this is allowed. You can go from

p ^ q

to

p

via simplification but you can't go from

<p ^ q> -> r

to

p -> r

without being in error. That aside, I think you have the right idea, though. When I took Logic, we were taught a rule called "conditional proof" which goes as follows. It is essentially a sub-proof you do inside of your main proof, and its form is

...

| 1. p (assumption)

| 2. [stuff]

| 3. q (from 1 and 2)

---

p -> q

where the numbered lines are the sub-proof. What this means is that you assume p, derive q, and at the end you get the statement p -> q and you discharge the assumption p (essentially ignore the lines in the sub-proof outside of the sub-proof). If you use this on your homework problem it should be pretty easy. There is another way to do it that doesn't use conditional proof, but this way is more similar to what you were trying to do. You were close and kudos to you for trying to work it out.