Lipschitz continuity of $f(x,y)=4x^2+xy-\frac{1}{y-1}$ on an open set $U \subset \mathbb{R} \times (\mathbb{R} \setminus \lbrace 1 \rbrace)$ 
Problem: Find an open set $U \subset \mathbb{R} \times (\mathbb{R} \setminus \lbrace 1 \rbrace )$ which includes the points $(0, 1/2$) and $(0,3/2)$ such that the function $f(x,y)=4x^2+xy-\frac{1}{y-1}$ is Lipschitz-continuous 

Doubts and concerns: Here $f: \mathbb{R}^2 \to \mathbb{R}$ is a function of two variables. In class we had the following definition  of Lipschitz continuity: 

Definition: A function $f: I \times U \to \mathbb{R}^m$ is relative to $x$ Lipschitz-continuous  $$ \iff \exists L \geq 0, \forall t \in I, \forall (x_1,x_2) \in U^2, \text{ s.t. } d(f(t,x_1),f(t,x_2)) \leq L d(x_1,x_2)$$



*

*So what bothers me is the notion of the relative to (variable). Would you say that $f$ is Lipschitz-continuous if it is Lipschitz-continuous relative to all variables?

*In class we've only dealt with real valued functions $f: \mathbb{R} \to \mathbb{R}$ which are indeed super easy to verify using the criteria of a bound derivative on the given set for example


My approach: After browsing the internet for a while I have found problems such as $f(t,x)$ being solved by partially differentiating with respect to the second variable. Applying this 'self-taught method' I get: $$ f(x,y)_y = x - \frac{1}{(y-1)^2} \tag*{(*)}$$
Looking at the given points $(x_1,y_1)=(0,1/2)$ and $(x_2,y_2)=(0,3/2)$ I could say that $x=0$ and $1/2 < y < 3/2 \implies 0< y <1$ 
which just runs me into more trouble, considering the equation (*)
I would appreciate some basic steps to get me started and maybe some clarification on the definition with the relative to/respect to that bothers me. 
 A: A function $f \,:\, \mathbb{R}^n \to \mathbb{R}^m$ is lipschitz continuous on $D \subset \mathbb{R}^n$ iff there is an $L \in \mathbb{R}$ such that $$
  \|f(x) - f(y)\| \leq L\|x-y\| \text{ for all $x,y \in D$.}
$$
Let's assume that $n > 1$, then we can also interpret such an $f$ as a function $$
  f \,:\, \mathbb{R}\times\mathbb{R}^{n-1} \to \mathbb{R}^m \,:\,
  (t,(x_1,\ldots,x_{n-1})) \mapsto f((t,x_0,\ldots,x_n)) \text{.}
$$
$f$ being lipschitz continuous relative to $t$ then is a weaker concept than (full) lischitz continuity of $f$ as function $\mathbb{R}^n \to \mathbb{R}^m$. It only tells us that for every fixed t, the function $x \mapsto f(t,x)$ is lipschitz continuous, and that one global bound $L$ works for all these functions. But it doesn't give us any bound for $\|f(t_1,x) - f(t_2,y)\|$ for $t_1 \neq t_2$.
However, if $f \,:\, \mathbb{R}^k\times \mathbb{R}^l \to \mathbb{R}^m \,:\, (x,y) \mapsto f(x,y)$ is lipschitz continuous relative to $x$ and to $y$ with lipschitz constants $L_x$ respectively $L_y$, then $f$ is lipschitz-continuous as a function $\mathbb{R}^{k+l} \to \mathbb{R}^m$ because $$\begin{eqnarray}
  \|f(a_x,a_y) - f(b_x,b_y)\| &=& \|f(a_x,a_y) - f(a_x,b_y) + f(a_x,b_y) - f(b_x,b_y)\| \\
  &\leq& \underbrace{\|f(a_x,a_y) - f(a_x,b_y)\|}_{\leq L_x\|a_y - b_y\|}
         + \underbrace{\|f(a_x,b_y) - f(b_x,b_y)\|}_{\leq L_y\|a_x - b_x\|} \\
  &\leq& L'\left( \|a_x - b_x\| + \|a_y - b_y\| \right) \quad L':=\max\{L_x,L_y\}\\
  &\leq& L'\left( \|(a_x,0) - (b_x,0)\| + \|(0,a_y) - (0,b_y)\|\right) \\
  &\leq& L'\sqrt{2} \|(a_x,a_y) - (b_x,b_y)\| \\
  &=&    L\|(a_x,a_y) - (b_x,b_y)\| \quad L:=\sqrt{2}L' = \sqrt{2}\max\{L_x,L_y\}  \text{.}
\end{eqnarray}$$
(The last line uses that $a + b \leq \sqrt{2}\sqrt{a^2 + b^2}$. Also, $(a,0)$ is to be read as $a\in\mathbb{R}^k$ with $l$ zeros appended to make it a vector in $\mathbb{R}^{k+l}$, and the same goes for $(0,b)$, only that here $k$ zeros are prepended).
In your case, you're dealing with (full) lipschitz continuity. It's easy to see that the only regions of $\mathbb{R}^2$ which cause trouble are those which either include $y$-values arbitrarily close to $1$, or which are unbounded.
