Given differential equation: $\frac{d}{dx}(xy^{2})=2y(1+x^{2})$

Differentiate the left side using the product rule: $y^{2} + x(2y\frac{dy}{dx}) = 2y(1+x^{2})$

Rearrange the equation: $y^{2} + 2xy\frac{dy}{dx} = 2y + 2yx^{2}$

Divide by $y$ (assuming $y \neq 0$): $y + 2x\frac{dy}{dx} = 2 + 2x^{2}$

Isolate the $\frac{dy}{dx}$ term: $2x\frac{dy}{dx} = 2 + 2x^{2} - y$

Rewrite the equation: $\frac{dy}{dx} = \frac{2 + 2x^{2} - y}{2x}$

This differential equation is not easily separable or linear. Let's try a different approach from the beginning.

Given: $\frac{d}{dx}(xy^{2})=2y(1+x^{2})$

Integrate both sides with respect to $x$: $\int \frac{d}{dx}(xy^{2}) dx = \int 2y(1+x^{2}) dx$

$xy^{2} = \int 2y(1+x^{2}) dx$

This approach requires us to express $y$ as a function of $x$ inside the integral, which is not straightforward. Let's go back to the differentiated form.

From $y^{2} + 2xy\frac{dy}{dx} = 2y(1+x^{2})$, we can rewrite it as:

$y^{2} + 2xy\frac{dy}{dx} = 2y + 2yx^{2}$

Divide throughout by $y$: $y + 2x\frac{dy}{dx} = 2 + 2x^{2}$

Rearrange: $2x\frac{dy}{dx} = 2x^{2} - y + 2$

Let's try another approach. Rewrite the original equation as $d(xy^2) = 2y(1+x^2)dx$.

Then, $xy^2 = \int 2y(1+x^2) dx$. This is difficult to solve directly.

Let's consider the original equation again: $\frac{d}{dx}(xy^{2})=2y(1+x^{2})$.

Expanding the derivative: $y^2 + 2xy \frac{dy}{dx} = 2y + 2yx^2$.

Dividing by $y$ (assuming $y \neq 0$): $y + 2x \frac{dy}{dx} = 2 + 2x^2$.

Rearranging: $2x \frac{dy}{dx} = 2x^2 - y + 2$.

This is not a separable equation in its current form.

Let's try to rearrange the equation as follows:

$\frac{d(xy^2)}{dx} = 2y(1+x^2)$

$d(xy^2) = 2y(1+x^2)dx$

This form is still difficult to integrate directly.

Let's go back to $y + 2x\frac{dy}{dx} = 2 + 2x^{2}$.

Rearranging, we get $2x\frac{dy}{dx} = 2x^2 - y + 2$.

This is a first-order differential equation, but it's not linear or separable.

Let's try to find an integrating factor. Rewrite the equation as:

$\frac{dy}{dx} + \frac{1}{2x}y = x + \frac{1}{x}$

The integrating factor is $e^{\int \frac{1}{2x} dx} = e^{\frac{1}{2} \ln x} = e^{\ln \sqrt{x}} = \sqrt{x}$.

Multiply the equation by the integrating factor:

$\sqrt{x} \frac{dy}{dx} + \frac{1}{2\sqrt{x}}y = x\sqrt{x} + \frac{1}{\sqrt{x}}$

$\frac{d}{dx}(y\sqrt{x}) = x\sqrt{x} + \frac{1}{\sqrt{x}}$

Integrate both sides with respect to $x$:

$\int \frac{d}{dx}(y\sqrt{x}) dx = \int (x\sqrt{x} + \frac{1}{\sqrt{x}}) dx$

$y\sqrt{x} = \int (x^{3/2} + x^{-1/2}) dx$

$y\sqrt{x} = \frac{x^{5/2}}{5/2} + \frac{x^{1/2}}{1/2} + C$

$y\sqrt{x} = \frac{2}{5}x^{5/2} + 2x^{1/2} + C$

Divide by $\sqrt{x}$:

$y = \frac{2}{5}x^2 + 2 + \frac{C}{\sqrt{x}}$