Answer
(a) This graph is symmetric about the $y$-axis, is not symmetric about the $x$-axis, and is not symmetric about the origin.
(b) This graph is not symmetric about the $y$-axis, is not symmetric about the $x$-axis, and is symmetric about the origin.
(c) This graph is symmetric about the $y$-axis, is symmetric about the $x$-axis, and is symmetric about the origin.
Work Step by Step
(a) For the graph of $x^4=2y^3+y$ we have
(i) $$x \to -x \quad \Rightarrow \quad (x^4=2y^3+y) \to ((-x)^4=x^4=2y^3+y),$$so it is symmetric about the $y$-axis,
(ii) $$y \to -y \quad \Rightarrow \quad (x^4=2y^3+y) \to (x^4=2(-y)^3-y=-2y^3-y),$$so it is not symmetric about the $x$-axis,
(iii) $$x \to -x, \quad y \to -y \quad \Rightarrow \quad (x^4=2y^3+y) \to ((-x)^4=x^4=2(-y)^3-y=-2y^3-y),$$so it is not symmetric about the origin.
(b) For the graph of $y= \frac{x}{3+x^2}$ we have
(i) $$x \to -x \quad \Rightarrow \quad (y=\frac{x}{3+x^2}) \to (y= \frac{-x}{3+(-x)^2}=-\frac{x}{3+x^2}),$$so it is not symmetric about the $y$-axis,
(ii) $$y \to -y \quad \Rightarrow \quad (y=\frac{x}{3+x^2}) \to (-y=\frac{x}{3+x^2}),$$so it is not symmetric about the $x$-axis,
(iii) $$x \to -x, \quad y \to -y \quad \Rightarrow \quad (y=\frac{x}{3+x^2}) \to (-y= \frac{-x}{3+(-x)^2} \quad \Rightarrow \quad y=\frac{x}{3+x^2}),$$so it is symmetric about the origin.
(c) For the graph of $y^2=|x|-5$ we have
(i) $$x \to -x \quad \Rightarrow \quad (y^2=|x|-5) \to (y^2=|-x|-5=|x|-5),$$ so it is symmetric about the $y$-axis,
(ii) $$y \to -y \quad \Rightarrow \quad (y^2=|x|-5) \to ((-y)^2=y^2=|x|-5),$$so it is symmetric about the $x$-axis,
(iii) $$x \to -x, \quad y \to -y \quad \Rightarrow \quad (y^2=|x|-5) \to ((-y)^2=y^2=|-x|-5=|x|-5),$$so it is symmetric about the origin.
