Technically speaking, the function $\displaystyle{f(x) = \frac{\sin x}{x}}$ is not Riemann integrable on $(0, \infty)$, but rather improperly Riemann integrable on $(0, \infty)$.
The construction of the Riemann integral only works for bounded intervals. We can extend this construction to unbounded intervals like $(0, \infty)$, but that requires an additional limiting process. It is the first construction (Riemann integrals for bounded integrals) that the Lebesgue integral generalizes.
What you write is more or less true. First note that it is not enough to assume that $f(x,t)$ is differentiable at $x_0$ (for (almost) all $t$), we need that there is a neighorhood $(x_0 - \varepsilon, x_0 + \varepsilon) =:I$ such that $f(\cdot, t)$ is differentiable in $x$ for all $x \in I$ and (almost) all $t \in A$, where the "exceptional null set" must not depend on $x$.
The assumption $|\partial f/\partial x (x,t)| \leq g(x)$ only makes sense in this setting, otherwise we could only require $x_0$ instead of $x$ and the proof also needs this assumption (see below).
Now, let $h_n \to 0$ with $h_n \neq 0$ for all $n$ and w.l.o.g. with $x_0 + h_n \in I$ for all $n$. Let
$$
F(x) := \int_A f(x,t) \, dt \text{ for } x \in I,\\
G(x) := \int_A \frac{\partial f}{\partial x}(x,t) \, dt \text{ for } x \in I.
$$
Here, the integral defining $F$ is interpreted as an improper Riemann integral (it is possible that this integral does not exist as a Lebesgue integral), but the integral defining $G$ is interpreted as a Lebesgue integral. If $t \mapsto \frac{\partial f}{\partial x}$ is Riemann integrable over every compact subinterval of $A$ (for example, if it is continuous in $t$), then it can also be interpreted as an improper Riemann integral, at least if we assume that the dominating function $g$ is (improper) Riemann integrable.
Observe that the assumptions guarantee that $F,G$ are well-defined. Measurability of $t \mapsto \frac{\partial f}{\partial x}(x,t)$ for each $x \in I$ is implied by the pointwise convergence $$\frac{\partial f}{\partial x}(x,t) = \lim_n \frac{f(x+h_n, t) - f(x,t)}{h_n},
$$
which will also be employed below.
We now have
\begin{eqnarray*}
\left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| & = & \left|\int_{A}\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\,{\rm d}t\right|\\
& \leq & \int_{A}\left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\right|\,{\rm d}t.
\end{eqnarray*}
Note that the integrands are measurable and that for (almost) all $t \in A$, the mean value theorem yields some $\xi_{n,t} \in I$ (even between $x_0$ and $x_0 +h_n$) such that
$$
\left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}\right|=\left|\frac{\partial f}{\partial x}\left(\xi_{n,t},t\right)\right|\leq g\left(t\right).
$$
This shows that the integrands in the integrals are Lebesgue-integrable and dominated by an integrable function. As the integrand in the last integral converges pointwise to $0$ for $n \to \infty$, the dominated convergence theorem yields
$$
\left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| \to 0,
$$
so that $F$ is differentiable (in $x_0$) with the expected derivative.
Best Answer
The function is not Lebesgue integrable.
Denote $A$ the integral of the absolute value of the function.Then $$ A=\int_{[-1,1]}\int_{[-1,1]}\frac{|x^2-y^2|}{(x^2+y^2)^2}dxdy$$ $$\geq \int_{B((0,1)}\frac{|x^2-y^2|}{(x^2+y^2)^2}dxdy$$
Now use polar coordinates and the fact that $\int_0^{2\pi}|\cos^2{x}-\sin^2{x}|dx=4$