Find the point where the curve $x+y=e^{xy}$ has vertical tangent.
That means $\frac{dx}{dy}=0$ at $(h,k)$
So differentiating given curve w.r.t $y$ both side we get
$$\frac{dx}{dy}+1=e^{xy}\left(x+y \frac{dx}{dy}\right)$$ putting $\frac{dx}{dy}=0$ we get
$$e^{hk}=\frac{1}{h} \tag{1} $$
Also $$h+k=e^{hk}=\frac{1}{h}$$
So $$k=\frac{1}{h}-h$$ substituting $k$ above in $(1)$ we get
$$e^{1-h^2}=\frac{1}{h}$$
that is $$eh= e^{h^2}$$
By guess $h=1$ but graph gives another solution , how can we reject that?
Simply put, you cannot reject the other solution, because it too yields a vertical tangent.
The numeric value of this solution is $$h \approx 0.45076365201730712954444197879707455059948938737236.$$
At this value, we find $$k \approx 1.7676938378993928727733449223485097251562623801697.$$
You may verify that $e^{hk} - 1/h \approx 0$ for this point, to the numerical precision given.
The graph of the relation $x + y = e^{xy}$ is shown below. Note the obvious symmetry:
The graph also suggests that a coordinate transformation could shed more light on the function's behavior; i.e., let us perform the rotation $$(u,v) = \left(\frac{x+y}{\sqrt{2}}, \frac{y-x}{\sqrt{2}}\right).$$ This transforms $(x,y) = (1,0)$ to $(u,v) = (1/\sqrt{2}, -1/\sqrt{2})$; i.e., it performs a clockwise rotation of $\pi/4$ about the origin. Then the implicit relation becomes $$e^{(u^2-v^2)/2} = u\sqrt{2},$$ which can be solved explicitly for $v$ to yield $$v = \pm g(u) = \pm \sqrt{u^2 - \log (2u^2)}, \quad u > 0.$$ This proves that the original curve has a slant asymptote where $x+y = 0$; it also shows that $g(u) \sim u$ as $u \to \infty$, since $u^2 \gg \log (2u^2)$. But since $g(1) = \sqrt{1 - \log 2} < 1$, it also shows that $g$ must approach $u$ from below: indeed, $$g'(u) = \frac{u^2-1}{u g(u)}$$ implies $g'(1) = 0$, and $g'(2) = 3/(2 \sqrt{4-3 \log 2}) > 1$, so by the intermediate value theorem, there is a $u \in (1,2)$ such that $g'(u) = 1$, corresponding to an $(x,y)$ for which $dx/dy = 0$.