Yes, the limit is $0$. You can also prove that using the fact that$$\frac{xy\sin y}{3x^2+y^2}=\frac{xy^2}{3x^2+y^2}\times\frac{\sin y}y,$$that$$\lim_{(x,y)\to(0,0)}\frac{xy^2}{3x^2+y^2}=0$$ and that $\lim_{y\to0}\frac{\sin y}y=1$.
It is meaningless state that $\lim\limits_{x\to \infty} \sqrt{x^2+x}-x \ne x - x$ and $\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = x + x$ we should state that
$$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x = \lim\limits_{x\to \infty} \frac1{2}+o(1/x)=\frac12$$
and
$$\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = \lim\limits_{x\to \infty} 2x+o(1)=\infty$$
the explanation in both case is in binomial first order approximation that is
$$\sqrt{x^2+x}=x\left(1+\frac1x\right)^\frac12= x\left(1+\frac1{2x}+o\left(\frac1x\right)\right)=x+ \frac1{2}+o\left(1\right) $$
which means that for $x$ large we have
$$\sqrt{x^2+x}\sim x+ \frac1{2}$$
and therefore
$$\sqrt{x^2+x}-x \sim \frac 12$$
$$\sqrt{x^2+x}+x \sim 2x+\frac 12$$
Edit
Note that for $\frac{x}{ \sqrt{x^2+x} + x}$ it is not correct to state that the denominator is $2x$ the complete steps are
$$\frac{x}{ \sqrt{x^2+x} + x}=\frac x x \frac{1}{ \sqrt{1+1/x} + 1} \to \frac12$$
For $\sqrt{x^2+x} - x$ indeed is not correct take$ \sqrt{x^2+x}=x$ what is true is that $\sqrt{x^2+x}\sim x+\frac12$.
If we use that approximation, it works with both limits. In this particular case first order approximation works and we can use it to evaluate both limits.
Best Answer
When we write
$$\lim_{x\to\infty}$$
we usually mean
$$\lim_{x\to+\infty}$$
when we need to be more clear in that the latter is preferable.
When we want indicate the two possiblities we can use
$$\lim_{|x|\to\infty}$$