The problem asks for the slope of the tangent to the curve ${\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}$ at the point $(1, 3)$. The slope of the tangent is given by the value of the derivative, $\frac{dy}{dx}$, at the given point.

Step 1: Verify the point is on the curve

First, let's check if the point $(1, 3)$ lies on the given curve by substituting $x=1$ and $y=3$ into the equation.

Left Hand Side (LHS): ${\left( {y - {x^5}} \right)^2} = {\left( {3 - {1^5}} \right)^2} = {\left( {3 - 1} \right)^2} = {2^2} = 4$ Right Hand Side (RHS): $x{\left( {1 + {x^2}} \right)^2} = 1{\left( {1 + {1^2}} \right)^2} = 1{\left( {1 + 1} \right)^2} = {2^2} = 4$ Since LHS = RHS, the point $(1, 3)$ is indeed on the curve.

Step 2: Differentiate the equation with respect to x

We will use implicit differentiation to find $\frac{dy}{dx}$. The given equation is: ${\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}$

Differentiating the LHS using the chain rule: $\frac{d}{dx} {\left( {y - {x^5}} \right)^2} = 2(y - x^5) \cdot \frac{d}{dx}(y - x^5)$ $= 2(y - x^5) \left( \frac{dy}{dx} - 5x^4 \right)$

Differentiating the RHS using the product rule and chain rule: Let $u = x$ and $v = (1 + x^2)^2$. Then $\frac{du}{dx} = 1$. $\frac{dv}{dx} = 2(1 + x^2) \cdot \frac{d}{dx}(1 + x^2) = 2(1 + x^2)(2x) = 4x(1 + x^2)$ So, the derivative of the RHS is: $\frac{d}{dx} \left[ x{\left( {1 + {x^2}} \right)^2} \right] = \frac{du}{dx}v + u\frac{dv}{dx}$ $= 1 \cdot (1 + x^2)^2 + x \cdot [4x(1 + x^2)]$ $= (1 + x^2)^2 + 4x^2(1 + x^2)$ Factoring out $(1 + x^2)$: $= (1 + x^2)[(1 + x^2) + 4x^2] = (1 + x^2)(1 + 5x^2)$

Step 3: Equate the derivatives

Now, we equate the derivatives of the LHS and RHS: $2(y - x^5) \left( \frac{dy}{dx} - 5x^4 \right) = (1 + x^2)(1 + 5x^2)$

Step 4: Substitute the point (1, 3) and solve for dy/dx

Substitute $x=1$ and $y=3$ into the differentiated equation: $2(3 - 1^5) \left( \frac{dy}{dx} - 5(1)^4 \right) = (1 + 1^2)(1 + 5(1)^2)$ $2(3 - 1) \left( \frac{dy}{dx} - 5 \right) = (1 + 1)(1 + 5)$ $2(2) \left( \frac{dy}{dx} - 5 \right) = (2)(6)$ $4 \left( \frac{dy}{dx} - 5 \right) = 12$

Now, we solve for $\frac{dy}{dx}$: $\frac{dy}{dx} - 5 = \frac{12}{4}$ $\frac{dy}{dx} - 5 = 3$ $\frac{dy}{dx} = 3 + 5$ $\frac{dy}{dx} = 8$

Thus, the slope of the tangent to the curve at the point $(1, 3)$ is 8.