The origin is a critical point of the function $f(x,y)=x^2+x\sin y+\log(1+y^2)$. Moreover, it is a relative minimum because of hessian matrix at that point is $\begin{pmatrix} 2 & 1\\ 1 &2\end{pmatrix}$.
The question is: is the origin an absolute minimum of $f$?
Since $f(0,0)=0$ we need to show that $f(x,y)\geq 0$. If we fix $y$ then $f(x,y)$ is a quadratic equation with discriminant $$D(y)=\sin^2(y)-4\log(1+y^2)$$ If we show that $D(y)\leq 0$ for every $y$ then it's clear that the quadratic $f(x,y)$ is always non-negative.
Clearly $D(y)$ is negative if $$4\log(1+y^2)>1\Leftrightarrow |y|>\sqrt{e^{1/4}-1}\simeq 0.3$$ If $0\leq y\leq\sqrt{e^{1/4}-1}$ we write $$D'(y)=\sin(2y)-\dfrac{8y}{1+y^2}=\dfrac{(y^2+1)\sin(2y)-8y}{y^2+1}$$ Now $\sin(2y)\leq 2y$ so that the numerator in $D'(y)$ is smaller than $$(y^2+1)2y-8y=2y(y^2-7)$$ which is negative if $0\leq y\leq\sqrt{e^{1/4}-1}$. Hence $D(y)$ is decreasing there and $D(y)\leq D(0)=0$. By symmetry this holds too if $\sqrt{e^{1/4}-1}\leq y\leq 0$ hence $D(y)\leq 0$ for every $y$.
For all fixed $y \in \Bbb{R}$ the minimum $$\min_{x \in \Bbb{R}} \{ x^2 + x\sin y + \log(1+y^2) \}$$ is achieved at $x= - \frac{1}{2} \sin y$. That's because $x^2 + x\sin y + \log(1+y^2)$ is a parabola with vertex at $x= - \frac{1}{2} \sin y$.
So, putting $x= - \frac{1}{2} \sin y$ you have that $$\min_{x,y \in \Bbb{R}} x^2 + x\sin y + \log(1+y^2) = \min_{y \in \Bbb{R}} \left( - \frac{1}{2} \sin y \right)^2 + \sin y \left( - \frac{1}{2} \sin y \right) + \log (1+y^2) =$$ $$= \min_{y \in \Bbb{R}} \log(1+y^2) -\frac{\sin^2 y}{4}$$
This brings you to study a one-variable function to conclude that the minimum is indeed $0$.
Going for the level curves: \begin{align} C &= f(x,y) = x^2+x \sin(y) + \ln(1+y^2) \\ &= (x + \sin(y)/2)^2 - (\sin(y)/2)^2 + \ln(1+y^2) \iff \\ x &= -\sin(y)/2 \pm \sqrt{C + (\sin(y)/2)^2- \ln(1+y^2)} \end{align} We have the condition $$ 0 \le C + (\sin(y)/2)^2- \ln(1+y^2) \iff \\ C \ge \ln(1+y^2) - (\sin(y)/2)^2 = - D(y) / 4 \ge 0 $$ where the definition and result for $D(y)$ by Olivier were used.
So we have only non-negative level curves, $C=0$ the one with the lowest value.
The image above shows some level curves with the one for $C=1$ in green.
Note: The vertical axis is the $x$-axis, the horizontal axis is the $y$-axis.
