Given the differential equation: $(1+x^{2})dy+(1-\tan^{-1}x)dx=0$.

We can rewrite it as: $\frac{dy}{dx} = -\frac{1-\tan^{-1}x}{1+x^{2}}$

Separating variables, we get: $dy = -\frac{1-\tan^{-1}x}{1+x^{2}}dx$

Integrating both sides:

$\int dy = -\int \frac{1-\tan^{-1}x}{1+x^{2}}dx$

Let $u = \tan^{-1}x$, then $du = \frac{1}{1+x^{2}}dx$

So, $\int dy = -\int (1-u)du$

$y = -\int (1-u)du = -u + \frac{u^{2}}{2} + C$

Substituting back $u = \tan^{-1}x$, we get: $y = -\tan^{-1}x + \frac{(\tan^{-1}x)^{2}}{2} + C$

Given the initial condition $y(0) = 1$, we can find the constant $C$:

$1 = -\tan^{-1}(0) + \frac{(\tan^{-1}(0))^{2}}{2} + C$

$1 = -0 + \frac{0^{2}}{2} + C$

$C = 1$

So, the solution is: $y(x) = -\tan^{-1}x + \frac{(\tan^{-1}x)^{2}}{2} + 1$

Now, we need to find $y(1)$:

$y(1) = -\tan^{-1}(1) + \frac{(\tan^{-1}(1))^{2}}{2} + 1$

Since $\tan^{-1}(1) = \frac{\pi}{4}$, we have:

$y(1) = -\frac{\pi}{4} + \frac{(\frac{\pi}{4})^{2}}{2} + 1$

$y(1) = -\frac{\pi}{4} + \frac{\frac{\pi^{2}}{16}}{2} + 1$

$y(1) = -\frac{\pi}{4} + \frac{\pi^{2}}{32} + 1$

$y(1) = \frac{\pi^{2}}{32} - \frac{\pi}{4} + 1$