Proving that the set of connectives complete is typically done by showing that they can express each of a known complete set such as {AND,NOT} or {NOR} -- only occasionally will it be easier to provide a direct procedure for expressing an arbitrary truth table.
On the other hand, proving a set of connectives is incomplete is usually done by an argument along the lines of:
All expressible functions of $n$ variables have such-and-such-property, because the set of functions with this property is closed under each of the connectives and the function that just returns one of the variable also has this property. However, this-or-that particular function of $n$ variables doesn't have such-and-such property, and therefore it cannot be expressed, Q.E.D.
The proof will be more or less elegant according to how simple a "such-and-such" property you can get away with. At worst, such-and-such property will be something like "is one of these 27 particular functions", and it will be quite tedious (but not difficult) to verify that the 27 given functions are indeed closed under each connective.
A systematic procedure when a set of connectives are given is to start by working out which functions of a single variable can be produced. If one of the four possible functions can't be expressed, there's your answer. Then look for which functions of two variables can be expressed -- start with $\{0,1,x,y,\bar x,\bar y\}$ and try all combinations of those with the given connectives. Either you will quickly reach a dead end (which gives you a small "such-and-such" property for the incompletness proof), or you reach a sufficiently nontrivial combination of $x$ and $y$ that you can produce NOR by combining it with appropriate negations, and then you have a complete set of connectives.
In your particular case, we quickly notice that $g(x,y)$ is the negation of either $x$ or $y$ (depending on how your notation works), and $h(x,x,x)$ is the constant function $0$. This takes care of all one-argument functions.
Now what happens if you let one of the inputs to $h$ be one of these constants, and the other two be free variables?
Any functionally complete set of connectives must have at least one connective $\$$ such that $\top\$\top$ is equal to $\bot$ and one connective such that $\bot\$\bot$ is equal to $\top$. That means you only need to check the $4$ connectives that satisfy this condition, depending on what they return for $\top\$\bot$ and $\bot\$\top$.
Best Answer
To prove that a set of connectives is functionally complete, you simply need to show that you can derive any other logical connective using only this restricted set.
In your case, you want to show that you can obtain definitions for $\left\{\land,\leftrightarrow,\neg\right\}$ from $\left\{\to,\lor\right\}$, i.e. the question you have to ask yourselves is:
And by definition, if you cannot show that the set $\left\{\to,\lor\right\}$ is functionally complete, then it is not.
Playing around with these questions should give you a good idea of how connectives are related to each other and what functional completeness really means.
For a more formal approach, consider the following characterization of functional completeness. A set of connectives is functionally complete if it is not:
Thus, if you can show that $\left\{\to,\lor\right\}$ has any of the above properties, then it is not functionally complete.