1. The problem is to find the partial derivative $\frac{\partial}{\partial x}$ of the function $y(1+y^2)$ where $y$ is implicitly defined by the equation $1,0 = \sqrt{x}$ (interpreted as $y = \sqrt{x}$).\n\n2. First, express $y$ explicitly: $y = \sqrt{x} = x^{\frac{1}{2}}$.\n\n3. The function to differentiate is $f(x) = y(1+y^2) = y + y^3$.\n\n4. Substitute $y = x^{\frac{1}{2}}$ into $f(x)$: $$f(x) = x^{\frac{1}{2}} + (x^{\frac{1}{2}})^3 = x^{\frac{1}{2}} + x^{\frac{3}{2}}.$$\n\n5. Differentiate $f(x)$ with respect to $x$ using the power rule $\frac{d}{dx} x^n = n x^{n-1}$:\n$$\frac{df}{dx} = \frac{1}{2} x^{-\frac{1}{2}} + \frac{3}{2} x^{\frac{1}{2}}.$$\n\n6. Simplify the derivative: $$\frac{df}{dx} = \frac{1}{2 \sqrt{x}} + \frac{3}{2} \sqrt{x}.$$\n\n7. Evaluate at $x=1$:\n$$\frac{df}{dx}\bigg|_{x=1} = \frac{1}{2 \cdot 1} + \frac{3}{2} \cdot 1 = \frac{1}{2} + \frac{3}{2} = 2.$$\n\nFinal answer: $$\frac{d}{dx} \left(y(1+y^2)\right) \bigg|_{x=1} = 2.$$