Express \(\(2)/((1+x)(1+3x)) ) in partial fractions. Hence solve the differential equation \(\(dy)/(dx) = (2y)/((1+x)(1+3x)) ), given that \(y=1 ) when \(x=0 ).

To find A, let $1+x=0 \implies x=-1$:

$$2 = A(1+3(-1)) + B(1-1)$$ $$2 = A(1-3) + B(0)$$ $$2 = -2A$$ $$A = -1$$

To find B, let $1+3x=0 \implies x=-\frac{1}{3}$:

$$2 = A\left(1+3\left(-\frac{1}{3}\right)\right) + B\left(1-\frac{1}{3}\right)$$ $$2 = A(1-1) + B\left(\frac{2}{3}\right)$$ $$2 = B\left(\frac{2}{3}\right)$$ $$B = 2 \times \frac{3}{2}$$ $$B = 3$$

$$\frac{2}{(1+x)(1+3x)} = \frac{-1}{1+x} + \frac{3}{1+3x}$$

The partial fraction decomposition is $\boxed{\frac{3}{1+3x} - \frac{1}{1+x}}$.

Part 2: Solve the Differential Equation

Given the differential equation $\frac{dy}{dx} = \frac{2y}{(1+x)(1+3x)}$.

$$\frac{1}{y} \frac{dy}{dx} = \frac{2}{(1+x)(1+3x)}$$ $$\int \frac{1}{y}\,dy = \int \frac{2}{(1+x)(1+3x)}\,dx$$

$$\ln|y| = \int \frac{3}{1+3x}\,dx - \int \frac{1}{1+x}\,dx$$

For $\int \frac{3}{1+3x}\,dx$, let $u = 1+3x$, $du = 3\,dx$.

$$\int \frac{1}{u}\,du = \ln|u| = \ln|1+3x|$$

For $\int \frac{1}{1+x}\,dx$, let $v = 1+x$, $dv = \,dx$.

$$\int \frac{1}{v}\,dv = \ln|v| = \ln|1+x|$$

So,

$$\ln|y| = \ln|1+3x| - \ln|1+x| + C$$

Using the logarithm property $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$:

$$\ln|y| = \ln\left|\frac{1+3x}{1+x}\right| + C$$

$$\ln|1| = \ln\left|\frac{1+3(0)}{1+0}\right| + C$$ $$0 = \ln\left|\frac{1}{1}\right| + C$$ $$0 = \ln(1) + C$$ $$0 = 0 + C$$ $$C = 0$$

$$\ln|y| = \ln\left|\frac{1+3x}{1+x}\right|$$

Exponentiate both sides to solve for $y$:

$$e^{\ln|y|} = e^{\ln\left|\frac{1+3x}{1+x}\right|}$$ $$|y| = \left|\frac{1+3x}{1+x}\right|$$

Since $y=1$ when $x=0$, $y$ is positive, so we can remove the absolute value signs for $y$. Assuming $1+3x$ and $1+x$ have the same sign around $x=0$:

$$y = \frac{1+3x}{1+x}$$

The solution to the differential equation is $\boxed{y = \frac{1+3x}{1+x}}$.