Your math is fine, although you might want to mention that you are invoking the fundamental theorem of arithmetic when you factor $x$, $y$, and $xy$ (uniquely) into primes.
Here's a general tip on proof-writing: I seldom come across a proof where I think "this would be clearer if the author used more symbols." Making your proof look more "mathy" is not something you need (or want) to strive for. On the other hand, it's important to write proofs in clear, correct English, with proper punctuation. You've done a good job of this.
It's not necessary to enumerate your steps. As you start writing longer proofs, you will find it tedious to do so. As long as your proof is organized neatly into correct sentences (and paragraphs, if the proof is longer), there's no need to number them.
You can prove it using "multiplicative" property and it's simple but I just solved it without it and I believe that it's much more beautiful.
Consider $$n=p_{1}^{a_1} p_2^{a_2} ... p_k^{a_k}$$
So $\Omega(n)=\sum a_i$ and $\lambda(n)=(-1)^{\Omega(n)}$
Now $$\sum_{d|n}\lambda(d)=\sum_{d|n}(-1)^{\Omega(d)}$$
So it's suffices to calculate $D$; the difference of the number of divisor with even $\Omega$ and odd $\Omega.$
Now consider $f(x)=(1+x+x^2+...+x^{a_1})(1+x+x^2+...+x^{a_2})...(1+x+x^2+...+x^{a_k}) $
The coefficient of $x^r$ in above expansion is equal to the number of solutions of this equation:
$$x_1+x_2+...+x_k=r $$
$$0 \le x_i \le a_i$$
which is the number of divisors of $n$ with $\Omega$ equals to $r$.
Hence, $f(-1)=D.$
But $f(-1)$ is $0$ if at least one of $a_i$ is odd and $f(-1)=1$ if $n$ is a perfect square. $Q.E.D$
As a second solution:
It's easy to check $\lambda=1_{Sq}*\mu $
$[$ Actually the summation has only one term.$]$
Now convolve both sides with the $1$ function [which is inverse of $\mu$]
$$ \lambda * 1=1_{Sq}*\mu*1=1_{Sq} $$
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Best Answer
Let the highest power of prime $p$ in $a,b,c$ be $A,B,C$ respectively.
As $ab$ is perfect square, $A+B$ is even
Similarly, $B+C$ will be even $\implies (A+B)+ (B+C)$ will be even
Now the highest power of $p$ in $ac$ will be $A+C$
Now as $A+C-(A+B+B+C)=2B$ is even, so will be $A+C$