Answer
$y'=\frac{2xy-1-x^4y^2}{2xy+2x^5y^3-x^2}$
Work Step by Step
By the implicit differentiating and the chain rule it follows:
$$(x^{2}y)'(\arctan)'(x^{2}y)=1+2xyy'$$
$$(x^{2}y)'\frac{1}{1+(x^{2}y)^{2}}=1+2xyy'$$
$$(2xy+x^2y')\frac{1}{1+(x^{2}y)^{2}}=1+2xyy'$$
$$\frac{2xy}{1+(x^{2}y)^{2}}+\frac{x^2y'}{1+(x^{2}y)^{2}}=1+2xyy'$$
$$\frac{2xy}{1+(x^{2}y)^{2}}-1=2xyy'-\frac{x^2y'}{1+(x^{2}y)^{2}}$$
$$\frac{2xy}{1+(x^{2}y)^{2}}-1=y'\left(2xy-\frac{x^2}{1+(x^{2}y)^{2}}\right)$$
$$y'=\frac{\frac{2xy}{1+(x^{2}y)^{2}}-1}{(2xy-\frac{x^2}{1+(x^{2}y)^{2}})}$$
$$y'=\frac{2xy-1-x^4y^2}{2xy+2x^5y^3-x^2}$$
