Domain: x^2+3>0, so D_f=\R. f is odd (f(-x)=-f(x)): the graph is symmetric about the origin.

Limits: \frac{x}{x^2+3}=\frac{1/x}{1+3/x^2}\to0 as x\to\pm\infty: y=0 is a horizontal asymptote.

First derivative: f'(x)=\frac{(x^2+3)-x\cdot2x}{(x^2+3)^2}=\frac{3-x^2}{(x^2+3)^2}. Zeros x=\pm\sqrt3. Sign: - on (-\infty,-\sqrt3), + on (-\sqrt3,\sqrt3), - on (\sqrt3,\infty). Local min at x=-\sqrt3, f(-\sqrt3)=-\frac{\sqrt3}{6}\approx-0.29; local max at x=\sqrt3, f(\sqrt3)=\frac{\sqrt3}{6}\approx0.29.

Second derivative: f''(x)=\frac{2x(x^2-9)}{(x^2+3)^3}=\frac{2x(x-3)(x+3)}{(x^2+3)^3}. Sign chart of x, x-3, x+3: f''<0 on (-\infty,-3), >0 on (-3,0), <0 on (0,3), >0 on (3,\infty). Concave on (-\infty,-3] and [0,3]; convex on [-3,0] and [3,\infty). Inflection points x=-3,0,3 with f=-\frac14,0,\frac14.

Global: f is decreasing on (-\infty,-\sqrt3] from 0, increasing to \sqrt3, then decreasing to 0. Since f\to0 at \pm\infty, \frac{\sqrt3}{6} is the global max value and -\frac{\sqrt3}{6} the global min value.

Sketch: from 0^- at -\infty down to (-\sqrt3,-0.29), up through (0,0) (steepest there, slope \frac13) to (\sqrt3,0.29), then down towards 0^+.