How to find the slope of $y.\ln x = x.\ln y$ at $x = e$? Let's say we have the following equation:-$$y.\ln x=x.\ln y$$



After graphing the equation on desmos (which included, not surprisingly, the line $y=x$), I realised that the equation has a slope of 1 at all points of the form $(a,a)$, except it had two slopes at $(e,e)$. I went on to try and algebraically find its second slope.
$$y.\ln x=x.\ln y$$
$$\ln x.y' + y/x = (x/y).y'+\ln y$$
$$y' = (\ln y-y/x)/(\ln x-x/y)$$
Thus I got an expression for $y'$ which becomes indeterminate at $x=e$. I wasn't sure how to apply a limit with two variables, and wasn't sure how I would get two slopes.
How do I approach finding both the slopes in this self-intersecting graph?
 A: The hint.
Show that if $x^y=y^x$, $x\neq y$, $x>1$ and $y>1$, so $x^y>e^e$, 
which says that $(e,e)$ is a needed point.  
Indeed, let $y=kx$, where $k>1$.
Thus, $$x^{kx}=(kx)^x$$ or
$$x=k^{\frac{1}{k-1}},$$ which gives $$y=k^{\frac{k}{k-1}}$$ and we need to prove that $f(k)>e^e,$ where
$$f(k)=k^\frac{k^{\frac{k}{k-1}}}{k-1}.$$
Now, $$f'(k)=\frac{f(k)\cdot k^\frac{k}{k-1}}{(k-1)^3}\cdot\left(\frac{k-1}{\sqrt k}+\ln k\right)\left(\frac{k-1}{\sqrt k}-\ln k\right)>0$$ because easy to show that $$\frac{k-1}{\sqrt k}-\ln k>0$$ for any $k>1$,
which says that $f$ increases and since
$$\lim_{k\rightarrow1^+}f(k)=e^e,$$ we are done.
We see that $$\inf_{k>1}f=e^e,$$ which says that an unique common point on the line $x^y=y^x,$ where $x\neq y$, $x>1$ and $y>1$ plus a point $(e,e)$ and on the line $y=x$ it's a point $(e,e).$
Now, easy to get our slope.
For $x\neq y$ by the Hospital's rule we obtain:
$$\lim_{k\rightarrow1^+}y'(x)=\lim_{k\rightarrow1^+}\frac{\ln{y}-\frac{y}{x}}{\ln{x}-\frac{x}{y}}=\lim_{k\rightarrow1+}\frac{\frac{k}{k-1}\ln{k}-k}{\frac{1}{k-1}\ln{k}-\frac{1}{k}}=$$
$$=\lim_{k\rightarrow1^+}\frac{k\ln{k}-k^2+k}{\ln{k}-1+\frac{1}{k}}=\lim_{k\rightarrow1^+}\frac{\ln{k}+1-2k+1}{\frac{1}{k}-\frac{1}{k^2}}=\lim_{k\rightarrow1^+}\frac{\frac{1}{k}-2}{-\frac{1}{k^2}+\frac{2}{k^3}}=-1.$$
A: Obviously one of the solutions is $x=y$ and the slope is $1$. For the other, observe that the curve is symmetric about $x=y$ and is smooth. Hence $-1$ because it must be orthogonal to the line.
A: From the equation of the curve, setting $x=y$ gives that $$x\log x=0,$$ which implies $x=0$ or $x=1.$ Thus, the only such point is $(1,1).$ And since you've got an expression for the slope of the curve at each point, it follows that at this point the slope is $$\frac{0-1}{0-1}=1.$$ Thus, there's no such point as $(e,e)$ on the curve.
As for the expression for the slope, that becomes undefined when $$\log x=x/y,$$ which is the same as requiring that $$y=\frac{x}{\log x}.$$ Now is there such a point on the curve? Substituting in the equation of the curve and simplifying gives $$\frac{\log x}{\log x}=\log\left(\frac{x}{\log x}\right),$$ whenever $x \ne 1.$ For the case when $x=1,$ we have from the discussion in paragraph the first that $y=1.$ But is it the case that $\log x=x/y$? Clearly this is false. So the slope does not exist only when $e\log x=x,$ or in other words when $$e^x=x^e.$$ The last equation has a root $x=e,$ so that $y=e.$ But this point is not on the curve. If this is the only solution of the equation $e^x=x^e,$ that means that the curve always has a slope at each of its points, which is suggested also by the graph displayed.
