PART $1$
We just need to apply the Implicit Function Theorem. Consider the mapping $F : \mathbb{R}^4 \to \mathbb{R}^2$ given by $F(x,y,u,v) = (\cos u + \sin v-x, \cos v + \sin u-y)$. The set of $(x,y,u,v) \in \mathbb{R}^4$ such that $F(x,y,u,v) = (0,0)$ is the surface you're interested in. Calculating the Jacobian Matrix of $F$:
$J_F = \left[ \begin{array}{cccc} -1 & 0 & -\sin u & \cos v \\ 0 & -1 & \cos u & -\sin v \end{array} \right] . $
Clearly the rank of $J_F$ is maximal, i.e. equal to two, for all $x,y,u$ and $v.$ It follows that $F$ has no critical values and that $(0,0)$ must be a regular value. This means that the surface is a smooth parametrisable surface in a neighbourhood of each of its points.
Moreover, at $(x,y,u,v) = (0,0,1,1)$ we see that the last two columns form an invertible two-by-two matrix. By the IFT, we can write $u$ and $v$ as smooth functions of $x$ and $y$. (Notice that this hold for all $u$ and $v$ with $\cos(u+v) \neq 0.$)
PART $2$
Assume that $u$ and $v$ are functions of $x$ and $y$, i.e. $u = u(x,y)$ and $v=v(x,y).$ Consider the equation $x = \cos u + \sin v.$ Differentiating with respect to $x$ gives $1 = -u_x\sin u + v_x\cos v.$ Differentiating with respect to $y$ gives $0 = -u_y\sin u + v_y\cos v.$ Applying a similar process to $y = \sin u + \cos v$ gives $0 = u_x\cos u -v_x\sin v$ and $1 = u_y\cos u - v_y\sin v.$ We get:
$\left[ \begin{array}{cc} -\sin u & \cos v \\ \cos u & -\sin v \end{array} \right]\left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y\end{array}\right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1\end{array} \right] $
Assuming that $\sin u \sin v - \cos u \cos v \neq 0$ we can rearrange this to give:
$\left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y\end{array}\right] = \frac{1}{\cos(u+v)}\left[ \begin{array}{cc} \sin v & \cos v \\ \cos u & \sin u \end{array} \right].$
It follows that:
$u_x^2+u_y^2 = v_x^2 + v_y^2 \iff \frac{\sin^2v}{\cos^2(u+v)} + \frac{\cos^2v}{\cos^2(u+v)} = \frac{\cos^2u}{\cos^2(u+v)} + \frac{\sin^2u}{\cos^2(u+v)} \, . $
Using the fact that $\cos^2\theta + \sin^2\theta = 1$ for all $\theta$, we see that this last statement is indeed true. (Assuming $\cos(u+v) \neq 0.$)