= Solution
First consider the usual local reduction with sufficiently smooth coefficients. The <hyperbolic partial differential equation> has two distinct characteristic directions because $S^2-RT>0$. Choose independent local <characteristic coordinates> $\xi,\eta$ whose gradients satisfy
$$
R\xi_x^2+2S\xi_x\xi_y+T\xi_y^2=0,\qquad
R\eta_x^2+2S\eta_x\eta_y+T\eta_y^2=0.
$$
Let $J=\xi_x\eta_y-\xi_y\eta_x\ne0$ and
$$
B=R\xi_x\eta_x+S(\xi_x\eta_y+\xi_y\eta_x)+T\xi_y\eta_y.
$$
The determinant of the transformed principal coefficient matrix gives $B^2=J^2(S^2-RT)>0$. By the <chain rule>, the left side of the original <partial differential equation> becomes
$$
2B\,u_{\xi\eta}+A_\xi u_\xi+A_\eta u_\eta,
\qquad A_\xi=R\xi_{xx}+2S\xi_{xy}+T\xi_{yy},
$$
with the analogous expression for $A_\eta$. The original first <derivatives> are $z_x=u_\xi\xi_x+u_\eta\eta_x$ and $z_y=u_\xi\xi_y+u_\eta\eta_y$. Substitution and division by $2B$ yield the requested mixed-derivative canonical form. The construction is local, in a neighborhood where the two characteristic foliations give a <diffeomorphism>.
There is a regularity qualification in the printed hypotheses: merely continuous coefficients need not give twice continuously differentiable <characteristic coordinates>. Such coordinates are needed for the second-derivative <chain rule> above. For example, take $R=0$, $S=1/2$ and $T=|x|$ near the origin. The characteristic equation factors as $\xi_y(\xi_x+|x|\xi_y)=0$. One family is $x=\text{constant}$ and the other has first integrals $\eta=h(y-x|x|/2)$. For an independent coordinate $h'\ne0$, but $\eta_{xx}$ has opposite one-sided limits at $x=0$. Hence no twice continuously differentiable characteristic coordinate system exists there. This illustrates that <continuous hyperbolic coefficients need not admit smooth characteristic coordinates>. The formal general argument requires additional smoothness, for example twice continuously differentiable coefficients. The concrete coefficients below are smooth, so this issue does not affect their solution.
For the concrete <partial differential equation>, put
$$
\xi=\frac{x}{y^2},\qquad\eta=y\quad(y>0).
$$
The <Jacobian determinant> is $1/y^2\ne0$. Direct use of the <chain rule> gives
$$
z_{xx}=\frac{u_{\xi\xi}}{y^4},\qquad
z_{xy}=-\frac{2\xi u_{\xi\xi}}{y^3}+\frac{u_{\xi\eta}}{y^2}-\frac{2u_\xi}{y^3}.
$$
Therefore
$$
\boxed{u_{\xi\eta}=\frac2\eta u_\xi.}
$$
To impose the data efficiently, let $v=z_x$. Then $2xv_x+yv_y=0$, whose <characteristic curves> preserve $x/y^2$, so $v=H(x/y^2)$. On the initial line $x=1$, the value $v(1,y)=y^2$ gives $H(\xi)=1/\xi$ for every $\xi>0$. Thus $z_x=y^2/x$, and integration in $x$ gives $z=y^2\log x+C(y)$. The remaining initial condition gives $C(y)=0$:
$$
\boxed{z(x,y)=y^2\log x,\qquad x>0,\ y>0.}
$$
Every point in this region is reached from the data by a characteristic with positive $x/y^2$. The line $x=0$ separates these characteristics from those with $x<0$, and the logarithmic divergence prevents a smooth extension of this solution through $x=0$. The original condition restricts us to $y>0$.
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