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Set $t=2^x>0$. The exponential function identity $4^x=(2^x{)}^2$ reduces the problem to the quadratic equationPositivity excludes $t=-1$, so $2^x=2$. Since the exponential function $2^x$ is strictly increasing, the unique answer is $\boxed{x=1}$.

Expanding the quadratic form givesMatching coefficients of $xy$ and $y^2$ yields

The chain rule gives $d(e^{xy})=e^{xy}(ydx+xdy)$, so the given total differential is exact. Thus $f=e^{xy}+C$; two such functions differ by a constant because their gradient difference vanishes on the connected plane. The assigned value at the origin forces $C=-1$, giving

The Newton root-finding iteration replaces the graph by its tangent line at $x_0$ and uses that line's zero:Comparing with the printed reciprocal-ratio form givesThis step requires $f^{\prime }(x_0)\ne 0$. If $f(x_0)=0$, the iteration has already reached a zero of a function and the printed reciprocal-ratio expression need not be evaluated.