Show that the equation $x^2y^2 +2e^{xy} -4 -2e^2 =0$ can be solved for $y$ in terms of $x$ in a neighborhood of the point $x=1$ with $y(1) =2.$ Calculate $\frac{\partial y}{\partial x}$ when $x=1$.
I know that I need to use the implicit function theorem. However, I'm not sure how to approach it since it gives us $y(1)=2$.
You are right in saying that you need to use implicit differentiation. Let us try to make sense of implicit differentiation then before we move on to solving the problem.
Consider $x^2y^2 = c$ at ($x = 1$, $y = 2$). (This is an easy problem for us to cross-check what we are doing is sensible.)
Let $f(x) = x^2 \\ g(y) = y^2$
This results in $\dfrac{df(x)}{dx} = 2x$ and $\dfrac{dg(y)}{dy} = 2y$.
By chain rule, $\dfrac{dg(y)}{dx} = \dfrac{dg(y)}{dy} \cdot \dfrac{dy}{dx} = 2y \cdot \dfrac{dy}{dx}$.
Now, we need to differentiate - $x^2\cdot y^2 = f(x)\cdot g(x) = c$, we invoke product rule for the same.
$\dfrac{d(f(x) \cdot g(y))}{dx} = 0 \\ \implies \dfrac{df(x)}{dx} \cdot g(y) + f(x) \cdot \dfrac{dg(y)}{dy} = 0\\ \implies 2xy^2 + 2x^2y\cdot \dfrac{dy}{dx} = 0$
We know the values of $x$ and $y$, so the only unknown in the above equation is $\dfrac{dy}{dx}$ which comes out to be -2.
Now, to the mentioned problem $x^2y^2 +2e^{xy} -4 -2e^2 =0$. We need to repeat similar steps as above for differentiating $e^{xy}$ and add to the result above.
You should get the resulting equation: $2xy^2 + 2x^2y\cdot \dfrac{dy}{dx} + e^{xy}(y + x\cdot \dfrac{dy}{dx}) = 0$
Once again, the only unknown is $\dfrac{dy}{dx}$, solving which you should get $-\dfrac{8 + 2e^2}{4 + e^2}$