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The proposed linear change of variables makes the two pairs of boundary lines constant-coordinate lines: $v=1,4$ and $u=0,7$. Thus the image of the parallelogram is the rectangle $0\le u\le 7$, $1\le v\le 4$.

The forward Jacobian matrix and its determinant areInverting gives $x=(u+3v)/7$, $y=(v-2u)/7$, so the area element is $dxdy=\frac{1}{7}dudv$. The absolute inverse Jacobian determinant, rather than the forward determinant, belongs in the transformed area integral. Consequentlywhich yields $\boxed{\frac{7}{2}\log 4=7\log 2}$. The transformed denominator is bounded below by one throughout the region, so no singular or improper integral is involved.

Work on a domain with $x\ne 0$, and suppose $f$ is differentiable at $u=y/x$. The multivariable chain rule givesMultiplying the first equation by $x$ and the second by $y$ cancels their common terms:The geometry is that $y/x$ stays fixed along a radial dilation $(x,y)\mapsto (\lambda x,\lambda y)$. Thus $w$ is constant along each admissible radial ray, and the differential operator $x{\partial }_x+y{\partial }_y$ differentiates along that ray. This is the degree-zero case of Euler homogeneous function theorem. The displayed quotient does not define a function at $x=0$ without additional extension data.

Start with the Maclaurin series $\cos x=1-x^2/2+x^4/24+O(x^6)$. Put $z=-x^2/2+x^4/24+O(x^6)$ into the binomial series $(1+z{)}^{-2}=1-2z+3z^2+O(z^3)$. Keeping terms through $x^4$ givessoSince $d(\tan x)/dx={\sec }^{2}x$ and $\tan 0=0$, term-by-term integration supplies the constant of integration and givesThe expansions hold near zero; the nearest real poles at $\pm \pi /2$ limit their radius of convergence to $\pi /2$. The derivative relation also independently checks all three displayed coefficients of the tangent function.