First of all it's a Lipschitz constant since the constant in Lipschitz continuity is not uniquely determined. In fact if $K$ is a Lipschitz constant then all values larger that $K$ are also Lipschitz constants.
Second $f$ is not Lipschitz continuous if you allow variation in all variables.
Now we can use the mean value theorem since $g(a)-g(b) = g'(\xi)(a-b)$ for some $\xi\in]a,b[$. This means that $|g(a)-g(b)| = |g'(\xi)||a-b|$. And if $g'(\xi)$ is bounded an upper bound of $g'(\xi)$ would be a Lipschitz constant.
For given values of two of the arguments the partial derivate of $f$ is bounded which is easy to see since one can easily find an upper bound of the derivate.
It's only in that sense $f$ is Lipschitz continuous. We can see that if you consider variation in all variables the derivate is not bounded and with unbounded derivates we can find secants with arbitrarily high slope which makes it non-Lipschitz continuous.
You know that $x_1=x_2$, and you also know that $f(x_1,y_1)=f(x_2,y_2)$. So, this means that $f(x_1,y_1)=f(x_1,y_2)$. Then
$$e^{x_1}cos(y_1) = e^{x_1}cos(y_2)$$
and also
$$e^{x_1}sin(y_1) = e^{x_1}sin(y_2)$$
Since $e^{x_1}>0$ for all values of $x_1$ from the definition of the exponential function, multiplying both sides of each equations above by $\frac{1}{e^{x_1}}$ yields
$$\begin{cases}cos(y_1) = cos(y_2) \\ sin(y_1) = sin(y_2)\end{cases}$$
But since we are considering $y\in ]0, 2\pi[$, if they have the same $cos$ and $sin$ simultaneously, then this implies that
$$y_1 = y_2$$
So, $(x_1,y_1) = (x_2, y_2)$, then $f$ is one-to-one in $A$.
OBS: You can actually check the last statement by checking the quadrants in the unit circle. You know that:
$$\begin{cases}cos(y_1) = cos(y_2) \\ sin(y_1) = sin(y_2)\end{cases}$$
And also that both $y_1,y_2\in]0,2\pi[$.Then, by analyzing all the cases:
If $sin(y_1)=sin(y_2) \geq 0$, if:
$cos(y_1)=cos(y_2) \geq 0$, then both $y_1,y_2\in]0,\frac{\pi}{2}]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
$cos(y_1)=cos(y_2) < 0$, then both $y_1,y_2\in]\frac{\pi}{2},\pi]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
If $sin(y_1)=sin(y_2) < 0$, if:
$cos(y_1)=cos(y_2) \geq 0$, then both $y_1,y_2\in[\frac{3\pi}{2},2\pi[$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
$cos(y_1)=cos(y_2) < 0$, then both $y_1,y_2\in]\pi,\frac{3\pi}{2}]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
Best Answer
It will be Lipschitz continuous on any bounded rectangle since each of the component functions $f_1(x,y)=x \cos (y)$ and $f_2=x \sin (y)$ of the vector-valued function $f=(f_1,f_2)$ are $C^1$.
In particular, here we have $|\nabla f_1|^2=\cos^2(y)+x^2\sin^2(y)$ and $|\nabla f_2|^2=\sin^2(y)+x^2\cos^2(y)$. Hence $|\nabla f_1|,|\nabla f_2| \leq 1+x^2$. By the mean value theorem and the Cauchy-Schwarz inequality, we thus have \begin{aligned}d\left( f(x,y), f(u,v) \right) &=\sqrt{\left(f_1(x,y)-f_1(u,v) \right)^2 +\left(f_2(x,y)-f_2(u,v) \right)^2 } \\& \leq |f_1(x,y)-f_1(u,v)|+|f_2(x,y)-f_2(u,v)| \\& \leq 2\sqrt{(x-u)^2+(y-v)^2}+2\sqrt{(x-u)^2+(y-v)^2} \\&=4\sqrt{(x-u)^2+(y-v)^2} \\&=4d\left( (x,y), (u,v) \right) \quad \text { for all } (x,y), (u,v) \in [0,1] \times \mathbb{R} \end{aligned}
Note: Since our bound on each of $\nabla f_1$ and $\nabla f_2$ depends only on the first variable, $f$ is Lipschitz on every $G \subseteq T \times \mathbb{R}$ whenever $T \subset \mathbb{R}$ is bounded.
Notice $f$ can't be bi-Lipschitz on any set on which it fails to be injective. In particular, for every $K \geq 1$ and any $x \in [0,1]$ we will have
$$\frac{1}{K}d\left((x, 0), (x,2\pi)\right)=\frac{2\pi}{K}\geq 0=d\left(f(x, 0), f(x,2\pi)\right).$$
Note: $f$ will fail to be injective on any set $S \subseteq \mathbb{R}^2$ where there exist either $(x_1, y_1), (-x_1,y_2 ) \in S$ such that $y_1 \equiv \pi +y_2\pmod {2\pi}$ or $(x_1, y_1), (x_1,y_2 ) \in S$ such that $y_1 \equiv y_2 \pmod {2\pi}$.