modus tollens in this example The statement "I'll have a party (P) if it's not raining (~R)" is represented as:
~R -> P
Then, by modus tollens, can I say, ~P -> ~(~R), meaning ~P -> R, essentially saying, "if I don't have a party, it is raining?" Isn't this a logical fallacy?
 A: From $\neg R \to P$ you can indeed infer $\neg P \to R$, thought this is typically considered Contraposition rather than Modus Tollens, which would infer $R$ from $\neg R \to P$ and $\neg P$
And no, this is not a fallacy. I think you might be thinking of the Denying the Antecedent Fallacy, which would try to infer $\neg P$ from $\neg R \to P$ and $R$ ... maybe it looks like you are dealing with this fallacy because of the juxtaposition of the two English sentences:
"I'll have a party if it is not raining"
and 
"If I don't have a party, it is raining"
... but note that the 'if' part (the antecedent) in the first sentence is the second half, whereas in the second sentence it is the first half.
A: Let's find a simpler example to work with so it's more apparent that modus tollens is indeed valid.
Assume $p \rightarrow q$ and $\neg q$ are true. By modus tollens, we may immediately conclude that $\neg p$ is true. 
To understand why, let's assume that $\neg p$ is false even though $p \rightarrow q$ and $\neg q$ are true. If $\neg p$ is false, this would mean $p$ is true. Then, by modus ponens, we could conclude that $q$ is true. But now we have a contradiction because we just stated $q$ and $\neg q$ are true. Hence, our assumption that $\neg p$ is false must be actually be false.
Therefore, if $p \rightarrow q$ and $\neg q$ are true, then $\neg p$ is true. Modus tollens is valid.
